Qwen2.5-Instruct-RLT-7B
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3 items
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Updated
Qwen2.5-7B-RLT-SFT-Stratos-warmup
このモデルは、RLT論文のSFT（Supervised Fine-Tuning）-warmup段階の成果です。本実験では、論文（Reinforcement Learning Teachers of Test Time Scaling）のリポジトリのコードと設定を使用しました。
本モデルは、単に上記リポジトリのコードを実行したものです。このモデルは、この段階におけるモデルの特性を学習し、把握することを目的としています。将来的には、この実験を32Bの規模に拡張する可能性があります。
本段階の目的
この段階の目的は、モデルが特定の固定された形式でテキストを出力できるようにすることです。
事前学習とファインチューニングの違い
事前学習（Pre-training）とファインチューニング（Fine-tuning）の主な違いは次のとおりです。
- 事前学習: すべてのトークンで損失（loss）を計算し、主に次のトークンを予測する方法を学習します。通常、チャットテンプレートはありません。
- ファインチューニング: 与えられたチャットテンプレートの下で、特定の入力に対して期待される回答を生成します。損失は期待される回答のトークンのみで計算され、目的は与えられた入力に対して期待される回答を生成する方法を学習することです。
PyTorchでは、デフォルトで損失を計算する必要のないトークンのラベルを-100に設定します。
学習データの例
このファインチューニングの学習過程で実際に使用されたデータと、プログラムが計算したトークンおよびそのラベルの例を見てみましょう。
<|im_start|>system\nYour role as an assistant involves providing precise and accurate solutions before providing detailed explanations with your full work showing your systematic thinking process leading to each solution. Your explanations should show how you engaged in a comprehensive cycle of analysis, summarizing, exploration, reassessment, reflection, backtracing, and iteration to develop well-considered thinking process. Please structure your response into two main sections: Solution and Explanation. In the Solution section, present your well-thought solution that accurately answers the question. The solution should remain a logical, accurate, concise expression style and detail necessary step needed to reach the conclusion, formatted as follows: <|begin_of_solution|> {final formatted, precise, and clear solution} <|end_of_solution|>. In the Explanation section, comprehensively detail your reasoning process using the specified format: <|begin_of_explanation|> {explanation with steps separated with '\\n\\n'} <|end_of_explanation|> Each step should show detailed considerations leading to your solutions such as analisying questions, summarizing relevant findings, brainstorming new ideas, verifying the accuracy of the current steps, refining any errors, and revisiting previous steps. <|im_end|>\n<|im_start|>user\nReturn your final response within \\boxed{}. Simplify $\\sqrt{25000}$.<|im_end|>\n<|im_start|>assistant\n<|begin_of_solution|>\n\nTo simplify , we start by factoring the number into a product of a perfect square and another number. \n\n1. **Prime Factorization**: \n . \n\n2. **Identify Perfect Squares**: \n The largest perfect square factors are (from ) and (from ). \n Multiplying these gives . \n\n3. **Rewrite the Square Root**: \n \\[\n \\sqrt{25000} = \\sqrt{2500 \\times 10} = \\sqrt{2500} \\times \\sqrt{10}\n \\] \n Since , this simplifies to: \n \\[\n 50 \\times \\sqrt{10}\n \\]\n\n4. **Verification**: \n - Alternative factorization: , leading to . \n - Numerical approximation confirms consistency.\n\nThus, the simplified form of is .\n\n<|end_of_solution|>\n\n<|begin_of_explanation|>\n\nOkay, so I need to simplify the square root of 25000. Let me think about how to approach this. Hmm, square roots can sometimes be simplified by factoring the number into a product of a perfect square and another number. That way, I can take the square root of the perfect square out of the radical. \n\nFirst, I should recall what a perfect square is. A perfect square is a number that is the square of an integer. Examples are 1, 4, 9, 16, 25, 36, etc. So, I need to find the largest perfect square that divides 25000. Let me start by breaking down 25000 into its prime factors. That might help me identify any perfect squares within it.\n\nStarting with 25000. Let's divide by 2 first. 25000 divided by 2 is 12500. Divide by 2 again gives 6250. One more time by 2 is 3125. Wait, 3125 is not divisible by 2. So, the number of 2s in the prime factorization is 3. Now moving on to 5. 3125 divided by 5 is 625. Divided by 5 again is 125. Again by 5 is 25, and once more by 5 is 5, and finally by 5 is 1. So, that's 5^5. \n\nPutting it all together, the prime factors of 25000 are 2^3 * 5^5. Now, to find the largest perfect square factor, I need to look at the exponents of the prime factors. A perfect square has even exponents in its prime factorization. So, for each prime factor, I can take as many pairs as possible.\n\nStarting with 2^3. The exponent is 3, which is odd. The largest even exponent less than 3 is 2, so we can take 2^2 as part of the perfect square. Then for 5^5, the exponent is 5. The largest even exponent less than 5 is 4, so we can take 5^4. Therefore, the largest perfect square factor is 2^2 * 5^4.\n\nCalculating that: 2^2 is 4, and 5^4 is 625. Multiplying 4 and 625 gives 2500. So, 25000 can be written as 2500 * 10. Let me check that: 2500 * 10 is indeed 25000. So, √25000 = √(2500 * 10). \n\nSince the square root of a product is the product of the square roots, this becomes √2500 * √10. Now, √2500 is a known value. 2500 is 50 squared because 50*50=2500. So, √2500 is 50. Therefore, √25000 = 50 * √10. \n\nBut wait, is 10 a prime number? Yes, 10 is 2*5, both primes. There are no perfect square factors in 10, so √10 cannot be simplified further. Therefore, the simplified form of √25000 is 50√10. \n\nAlternatively, maybe there's another way to approach this. Let me verify. 25000 is 25 * 1000. Because 25 times 1000 is 25000. Then, √25000 = √(25 * 1000) = √25 * √1000. √25 is 5, so that becomes 5√1000. But √1000 can be simplified further. 1000 is 100 * 10, so √1000 = √(100*10) = √100 * √10 = 10√10. Therefore, 5√1000 becomes 5*10√10 = 50√10. So, same result. \n\nAnother way: Maybe using exponent rules. Since 25000 is 25 * 10^3. So, 25 is 5^2 and 10^3 is (2*5)^3 = 2^3*5^3. So, altogether, 5^2 * 2^3 *5^3 = 2^3*5^(2+3) = 2^3*5^5. Which is the same prime factors as before. Then, when taking the square root, we can write it as 2^(3/2)*5^(5/2) = 2^(1 + 1/2)*5^(2 + 1/2) = 2*5^2 * √(2*5) = 2*25*√10 = 50√10. Yep, same answer. \n\nLet me also compute the numerical value to check. √25000. 25000 is 25*1000, so square root is 5*sqrt(1000). sqrt(1000) is about 31.6227766, so 5*31.6227766 is approximately 158.113883. Alternatively, 50*sqrt(10). sqrt(10) is approximately 3.16227766, so 50*3.16227766 is also 158.113883. So both ways, the numerical value is the same. \n\nTherefore, the simplified radical form is 50√10, and the numerical value is approximately 158.11. But the question asks to simplify, so the exact form is needed, not the decimal approximation. So the answer is 50√10. \n\nWait, but is there a way to write it without a square root? If 25000 is a perfect square itself. Let's check. What number squared is 25000? Let's see. 150^2 is 22500, 160^2 is 25600. So 25000 is between 150^2 and 160^2. 158^2 is 24964, which is close but not 25000. 159^2 is 25281, which is over. So, 25000 is not a perfect square. Therefore, the simplified form is indeed 50√10. \n\nAlternatively, if the original number was 2500, that's 50^2, but 25000 is 100 times larger. So, sqrt(25000) = sqrt(2500 * 10) = sqrt(2500) * sqrt(10) = 50*sqrt(10). Yep, same thing. \n\nSo, after all these methods and checks, I'm confident that the simplified form of √25000 is 50√10.\n\n<|end_of_explanation|><|im_end|>\n
Hugging Faceでは、どうしても上記の数式がレンダリングされてしまうため、以上のテキストについては、raw形式のREADMEファイルを参照してください。
これをトークナイズし、ファインチューニング過程でのモデルのトークンと対応するラベルを表示します。
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' by' 553
' ' 220
'2' 17
' first' 1156
'.' 13
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' divided' 17779
' by' 553
' ' 220
'2' 17
' is' 374
' ' 220
'1' 16
'2' 17
'5' 20
'0' 15
'0' 15
'.' 13
' Divide' 62902
' by' 553
' ' 220
'2' 17
' again' 1549
' gives' 6696
' ' 220
'6' 21
'2' 17
'5' 20
'0' 15
'.' 13
' One' 3776
' more' 803
' time' 882
' by' 553
' ' 220
'2' 17
' is' 374
' ' 220
'3' 18
'1' 16
'2' 17
'5' 20
'.' 13
' Wait' 13824
',' 11
' ' 220
'3' 18
'1' 16
'2' 17
'5' 20
' is' 374
' not' 537
' divisible' 74916
' by' 553
' ' 220
'2' 17
'.' 13
' So' 2055
',' 11
' the' 279
' number' 1372
' of' 315
' ' 220
'2' 17
's' 82
' in' 304
' the' 279
' prime' 10250
' factor' 8168
'ization' 2022
' is' 374
' ' 220
'3' 18
'.' 13
' Now' 4695
' moving' 7218
' on' 389
' to' 311
' ' 220
'5' 20
'.' 13
' ' 220
'3' 18
'1' 16
'2' 17
'5' 20
' divided' 17779
' by' 553
' ' 220
'5' 20
' is' 374
' ' 220
'6' 21
'2' 17
'5' 20
'.' 13
' Div' 8765
'ided' 4490
' by' 553
' ' 220
'5' 20
' again' 1549
' is' 374
' ' 220
'1' 16
'2' 17
'5' 20
'.' 13
' Again' 13759
' by' 553
' ' 220
'5' 20
' is' 374
' ' 220
'2' 17
'5' 20
',' 11
' and' 323
' once' 3055
' more' 803
' by' 553
' ' 220
'5' 20
' is' 374
' ' 220
'5' 20
',' 11
' and' 323
' finally' 5499
' by' 553
' ' 220
'5' 20
' is' 374
' ' 220
'1' 16
'.' 13
' So' 2055
',' 11
' that' 429
"'s" 594
' ' 220
'5' 20
'^' 61
'5' 20
'.' 13
' \n\n' 4710
'Putting' 97904
' it' 432
' all' 678
' together' 3786
',' 11
' the' 279
' prime' 10250
' factors' 9363
' of' 315
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' are' 525
' ' 220
'2' 17
'^' 61
'3' 18
' *' 353
' ' 220
'5' 20
'^' 61
'5' 20
'.' 13
' Now' 4695
',' 11
' to' 311
' find' 1477
' the' 279
' largest' 7772
' perfect' 4727
' square' 9334
' factor' 8168
',' 11
' I' 358
' need' 1184
' to' 311
' look' 1401
' at' 518
' the' 279
' ex' 505
'ponents' 2700
' of' 315
' the' 279
' prime' 10250
' factors' 9363
'.' 13
' A' 362
' perfect' 4727
' square' 9334
' has' 702
' even' 1496
' ex' 505
'ponents' 2700
' in' 304
' its' 1181
' prime' 10250
' factor' 8168
'ization' 2022
'.' 13
' So' 2055
',' 11
' for' 369
' each' 1817
' prime' 10250
' factor' 8168
',' 11
' I' 358
' can' 646
' take' 1896
' as' 438
' many' 1657
' pairs' 13530
' as' 438
' possible' 3204
'.\n\n' 382
'Starting' 24617
' with' 448
' ' 220
'2' 17
'^' 61
'3' 18
'.' 13
' The' 576
' exponent' 27690
' is' 374
' ' 220
'3' 18
',' 11
' which' 892
' is' 374
' odd' 10322
'.' 13
' The' 576
' largest' 7772
' even' 1496
' exponent' 27690
' less' 2686
' than' 1091
' ' 220
'3' 18
' is' 374
' ' 220
'2' 17
',' 11
' so' 773
' we' 582
' can' 646
' take' 1896
' ' 220
'2' 17
'^' 61
'2' 17
' as' 438
' part' 949
' of' 315
' the' 279
' perfect' 4727
' square' 9334
'.' 13
' Then' 5005
' for' 369
' ' 220
'5' 20
'^' 61
'5' 20
',' 11
' the' 279
' exponent' 27690
' is' 374
' ' 220
'5' 20
'.' 13
' The' 576
' largest' 7772
' even' 1496
' exponent' 27690
' less' 2686
' than' 1091
' ' 220
'5' 20
' is' 374
' ' 220
'4' 19
',' 11
' so' 773
' we' 582
' can' 646
' take' 1896
' ' 220
'5' 20
'^' 61
'4' 19
'.' 13
' Therefore' 15277
',' 11
' the' 279
' largest' 7772
' perfect' 4727
' square' 9334
' factor' 8168
' is' 374
' ' 220
'2' 17
'^' 61
'2' 17
' *' 353
' ' 220
'5' 20
'^' 61
'4' 19
'.\n\n' 382
'Calcul' 57908
'ating' 1095
' that' 429
':' 25
' ' 220
'2' 17
'^' 61
'2' 17
' is' 374
' ' 220
'4' 19
',' 11
' and' 323
' ' 220
'5' 20
'^' 61
'4' 19
' is' 374
' ' 220
'6' 21
'2' 17
'5' 20
'.' 13
' Multip' 58712
'lying' 6711
' ' 220
'4' 19
' and' 323
' ' 220
'6' 21
'2' 17
'5' 20
' gives' 6696
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'.' 13
' So' 2055
',' 11
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' can' 646
' be' 387
' written' 5326
' as' 438
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
' *' 353
' ' 220
'1' 16
'0' 15
'.' 13
' Let' 6771
' me' 752
' check' 1779
' that' 429
':' 25
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
' *' 353
' ' 220
'1' 16
'0' 15
' is' 374
' indeed' 12824
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
'.' 13
' So' 2055
',' 11
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' =' 284
' �' 11995
'�' 248
'(' 7
'2' 17
'5' 20
'0' 15
'0' 15
' *' 353
' ' 220
'1' 16
'0' 15
').' 568
' \n\n' 4710
'Since' 12549
' the' 279
' square' 9334
' root' 3704
' of' 315
' a' 264
' product' 1985
' is' 374
' the' 279
' product' 1985
' of' 315
' the' 279
' square' 9334
' roots' 19703
',' 11
' this' 419
' becomes' 9044
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
' *' 353
' �' 11995
'�' 248
'1' 16
'0' 15
'.' 13
' Now' 4695
',' 11
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
' is' 374
' a' 264
' known' 3881
' value' 897
'.' 13
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
' is' 374
' ' 220
'5' 20
'0' 15
' squared' 52263
' because' 1576
' ' 220
'5' 20
'0' 15
'*' 9
'5' 20
'0' 15
'=' 28
'2' 17
'5' 20
'0' 15
'0' 15
'.' 13
' So' 2055
',' 11
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
' is' 374
' ' 220
'5' 20
'0' 15
'.' 13
' Therefore' 15277
',' 11
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' =' 284
' ' 220
'5' 20
'0' 15
' *' 353
' �' 11995
'�' 248
'1' 16
'0' 15
'.' 13
' \n\n' 4710
'But' 3983
' wait' 3783
',' 11
' is' 374
' ' 220
'1' 16
'0' 15
' a' 264
' prime' 10250
' number' 1372
'?' 30
' Yes' 7414
',' 11
' ' 220
'1' 16
'0' 15
' is' 374
' ' 220
'2' 17
'*' 9
'5' 20
',' 11
' both' 2176
' primes' 49433
'.' 13
' There' 2619
' are' 525
' no' 902
' perfect' 4727
' square' 9334
' factors' 9363
' in' 304
' ' 220
'1' 16
'0' 15
',' 11
' so' 773
' �' 11995
'�' 248
'1' 16
'0' 15
' cannot' 4157
' be' 387
' simplified' 43799
' further' 4623
'.' 13
' Therefore' 15277
',' 11
' the' 279
' simplified' 43799
' form' 1352
' of' 315
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' is' 374
' ' 220
'5' 20
'0' 15
'√' 144336
'1' 16
'0' 15
'.' 13
' \n\n' 4710
'Alternatively' 92014
',' 11
' maybe' 7196
' there' 1052
"'s" 594
' another' 2441
' way' 1616
' to' 311
' approach' 5486
' this' 419
'.' 13
' Let' 6771
' me' 752
' verify' 10146
'.' 13
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' is' 374
' ' 220
'2' 17
'5' 20
' *' 353
' ' 220
'1' 16
'0' 15
'0' 15
'0' 15
'.' 13
' Because' 9211
' ' 220
'2' 17
'5' 20
' times' 3039
' ' 220
'1' 16
'0' 15
'0' 15
'0' 15
' is' 374
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
'.' 13
' Then' 5005
',' 11
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' =' 284
' �' 11995
'�' 248
'(' 7
'2' 17
'5' 20
' *' 353
' ' 220
'1' 16
'0' 15
'0' 15
'0' 15
')' 8
' =' 284
' �' 11995
'�' 248
'2' 17
'5' 20
' *' 353
' �' 11995
'�' 248
'1' 16
'0' 15
'0' 15
'0' 15
'.' 13
' �' 11995
'�' 248
'2' 17
'5' 20
' is' 374
' ' 220
'5' 20
',' 11
' so' 773
' that' 429
' becomes' 9044
' ' 220
'5' 20
'√' 144336
'1' 16
'0' 15
'0' 15
'0' 15
'.' 13
' But' 1988
' �' 11995
'�' 248
'1' 16
'0' 15
'0' 15
'0' 15
' can' 646
' be' 387
' simplified' 43799
' further' 4623
'.' 13
' ' 220
'1' 16
'0' 15
'0' 15
'0' 15
' is' 374
' ' 220
'1' 16
'0' 15
'0' 15
' *' 353
' ' 220
'1' 16
'0' 15
',' 11
' so' 773
' �' 11995
'�' 248
'1' 16
'0' 15
'0' 15
'0' 15
' =' 284
' �' 11995
'�' 248
'(' 7
'1' 16
'0' 15
'0' 15
'*' 9
'1' 16
'0' 15
')' 8
' =' 284
' �' 11995
'�' 248
'1' 16
'0' 15
'0' 15
' *' 353
' �' 11995
'�' 248
'1' 16
'0' 15
' =' 284
' ' 220
'1' 16
'0' 15
'√' 144336
'1' 16
'0' 15
'.' 13
' Therefore' 15277
',' 11
' ' 220
'5' 20
'√' 144336
'1' 16
'0' 15
'0' 15
'0' 15
' becomes' 9044
' ' 220
'5' 20
'*' 9
'1' 16
'0' 15
'√' 144336
'1' 16
'0' 15
' =' 284
' ' 220
'5' 20
'0' 15
'√' 144336
'1' 16
'0' 15
'.' 13
' So' 2055
',' 11
' same' 1852
' result' 1102
'.' 13
' \n\n' 4710
'Another' 14037
' way' 1616
':' 25
' Maybe' 10696
' using' 1667
' exponent' 27690
' rules' 5601
'.' 13
' Since' 8704
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' is' 374
' ' 220
'2' 17
'5' 20
' *' 353
' ' 220
'1' 16
'0' 15
'^' 61
'3' 18
'.' 13
' So' 2055
',' 11
' ' 220
'2' 17
'5' 20
' is' 374
' ' 220
'5' 20
'^' 61
'2' 17
' and' 323
' ' 220
'1' 16
'0' 15
'^' 61
'3' 18
' is' 374
' (' 320
'2' 17
'*' 9
'5' 20
')^' 29776
'3' 18
' =' 284
' ' 220
'2' 17
'^' 61
'3' 18
'*' 9
'5' 20
'^' 61
'3' 18
'.' 13
' So' 2055
',' 11
' altogether' 30055
',' 11
' ' 220
'5' 20
'^' 61
'2' 17
' *' 353
' ' 220
'2' 17
'^' 61
'3' 18
' *' 353
'5' 20
'^' 61
'3' 18
' =' 284
' ' 220
'2' 17
'^' 61
'3' 18
'*' 9
'5' 20
'^(' 13268
'2' 17
'+' 10
'3' 18
')' 8
' =' 284
' ' 220
'2' 17
'^' 61
'3' 18
'*' 9
'5' 20
'^' 61
'5' 20
'.' 13
' Which' 15920
' is' 374
' the' 279
' same' 1852
' prime' 10250
' factors' 9363
' as' 438
' before' 1573
'.' 13
' Then' 5005
',' 11
' when' 979
' taking' 4633
' the' 279
' square' 9334
' root' 3704
',' 11
' we' 582
' can' 646
' write' 3270
' it' 432
' as' 438
' ' 220
'2' 17
'^(' 13268
'3' 18
'/' 14
'2' 17
')*' 4806
'5' 20
'^(' 13268
'5' 20
'/' 14
'2' 17
')' 8
' =' 284
' ' 220
'2' 17
'^(' 13268
'1' 16
' +' 488
' ' 220
'1' 16
'/' 14
'2' 17
')*' 4806
'5' 20
'^(' 13268
'2' 17
' +' 488
' ' 220
'1' 16
'/' 14
'2' 17
')' 8
' =' 284
' ' 220
'2' 17
'*' 9
'5' 20
'^' 61
'2' 17
' *' 353
' �' 11995
'�' 248
'(' 7
'2' 17
'*' 9
'5' 20
')' 8
' =' 284
' ' 220
'2' 17
'*' 9
'2' 17
'5' 20
'*' 9
'√' 144336
'1' 16
'0' 15
' =' 284
' ' 220
'5' 20
'0' 15
'√' 144336
'1' 16
'0' 15
'.' 13
' Yep' 84194
',' 11
' same' 1852
' answer' 4226
'.' 13
' \n\n' 4710
'Let' 10061
' me' 752
' also' 1083
' compute' 12564
' the' 279
' numerical' 34776
' value' 897
' to' 311
' check' 1779
'.' 13
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
'.' 13
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' is' 374
' ' 220
'2' 17
'5' 20
'*' 9
'1' 16
'0' 15
'0' 15
'0' 15
',' 11
' so' 773
' square' 9334
' root' 3704
' is' 374
' ' 220
'5' 20
'*' 9
'sqrt' 26888
'(' 7
'1' 16
'0' 15
'0' 15
'0' 15
').' 568
' sqrt' 17946
'(' 7
'1' 16
'0' 15
'0' 15
'0' 15
')' 8
' is' 374
' about' 911
' ' 220
'3' 18
'1' 16
'.' 13
'6' 21
'2' 17
'2' 17
'7' 22
'7' 22
'6' 21
'6' 21
',' 11
' so' 773
' ' 220
'5' 20
'*' 9
'3' 18
'1' 16
'.' 13
'6' 21
'2' 17
'2' 17
'7' 22
'7' 22
'6' 21
'6' 21
' is' 374
' approximately' 13187
' ' 220
'1' 16
'5' 20
'8' 23
'.' 13
'1' 16
'1' 16
'3' 18
'8' 23
'8' 23
'3' 18
'.' 13
' Alternatively' 38478
',' 11
' ' 220
'5' 20
'0' 15
'*' 9
'sqrt' 26888
'(' 7
'1' 16
'0' 15
').' 568
' sqrt' 17946
'(' 7
'1' 16
'0' 15
')' 8
' is' 374
' approximately' 13187
' ' 220
'3' 18
'.' 13
'1' 16
'6' 21
'2' 17
'2' 17
'7' 22
'7' 22
'6' 21
'6' 21
',' 11
' so' 773
' ' 220
'5' 20
'0' 15
'*' 9
'3' 18
'.' 13
'1' 16
'6' 21
'2' 17
'2' 17
'7' 22
'7' 22
'6' 21
'6' 21
' is' 374
' also' 1083
' ' 220
'1' 16
'5' 20
'8' 23
'.' 13
'1' 16
'1' 16
'3' 18
'8' 23
'8' 23
'3' 18
'.' 13
' So' 2055
' both' 2176
' ways' 5510
',' 11
' the' 279
' numerical' 34776
' value' 897
' is' 374
' the' 279
' same' 1852
'.' 13
' \n\n' 4710
'Therefore' 54815
',' 11
' the' 279
' simplified' 43799
' radical' 17855
' form' 1352
' is' 374
' ' 220
'5' 20
'0' 15
'√' 144336
'1' 16
'0' 15
',' 11
' and' 323
' the' 279
' numerical' 34776
' value' 897
' is' 374
' approximately' 13187
' ' 220
'1' 16
'5' 20
'8' 23
'.' 13
'1' 16
'1' 16
'.' 13
' But' 1988
' the' 279
' question' 3405
' asks' 17064
' to' 311
' simplify' 39721
',' 11
' so' 773
' the' 279
' exact' 4734
' form' 1352
' is' 374
' needed' 4362
',' 11
' not' 537
' the' 279
' decimal' 12122
' approximation' 56204
'.' 13
' So' 2055
' the' 279
' answer' 4226
' is' 374
' ' 220
'5' 20
'0' 15
'√' 144336
'1' 16
'0' 15
'.' 13
' \n\n' 4710
'Wait' 14190
',' 11
' but' 714
' is' 374
' there' 1052
' a' 264
' way' 1616
' to' 311
' write' 3270
' it' 432
' without' 2041
' a' 264
' square' 9334
' root' 3704
'?' 30
' If' 1416
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' is' 374
' a' 264
' perfect' 4727
' square' 9334
' itself' 5086
'.' 13
' Let' 6771
"'s" 594
' check' 1779
'.' 13
' What' 3555
' number' 1372
' squared' 52263
' is' 374
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
'?' 30
' Let' 6771
"'s" 594
' see' 1490
'.' 13
' ' 220
'1' 16
'5' 20
'0' 15
'^' 61
'2' 17
' is' 374
' ' 220
'2' 17
'2' 17
'5' 20
'0' 15
'0' 15
',' 11
' ' 220
'1' 16
'6' 21
'0' 15
'^' 61
'2' 17
' is' 374
' ' 220
'2' 17
'5' 20
'6' 21
'0' 15
'0' 15
'.' 13
' So' 2055
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' is' 374
' between' 1948
' ' 220
'1' 16
'5' 20
'0' 15
'^' 61
'2' 17
' and' 323
' ' 220
'1' 16
'6' 21
'0' 15
'^' 61
'2' 17
'.' 13
' ' 220
'1' 16
'5' 20
'8' 23
'^' 61
'2' 17
' is' 374
' ' 220
'2' 17
'4' 19
'9' 24
'6' 21
'4' 19
',' 11
' which' 892
' is' 374
' close' 3265
' but' 714
' not' 537
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
'.' 13
' ' 220
'1' 16
'5' 20
'9' 24
'^' 61
'2' 17
' is' 374
' ' 220
'2' 17
'5' 20
'2' 17
'8' 23
'1' 16
',' 11
' which' 892
' is' 374
' over' 916
'.' 13
' So' 2055
',' 11
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' is' 374
' not' 537
' a' 264
' perfect' 4727
' square' 9334
'.' 13
' Therefore' 15277
',' 11
' the' 279
' simplified' 43799
' form' 1352
' is' 374
' indeed' 12824
' ' 220
'5' 20
'0' 15
'√' 144336
'1' 16
'0' 15
'.' 13
' \n\n' 4710
'Alternatively' 92014
',' 11
' if' 421
' the' 279
' original' 4024
' number' 1372
' was' 572
' ' 220
'2' 17
'5' 20
'0' 15
'0' 15
',' 11
' that' 429
"'s" 594
' ' 220
'5' 20
'0' 15
'^' 61
'2' 17
',' 11
' but' 714
' ' 220
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'0' 15
'0' 15
'0' 15
' is' 374
' ' 220
'1' 16
'0' 15
'0' 15
' times' 3039
' larger' 8131
'.' 13
' So' 2055
',' 11
' sqrt' 17946
'(' 7
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
')' 8
' =' 284
' sqrt' 17946
'(' 7
'2' 17
'5' 20
'0' 15
'0' 15
' *' 353
' ' 220
'1' 16
'0' 15
')' 8
' =' 284
' sqrt' 17946
'(' 7
'2' 17
'5' 20
'0' 15
'0' 15
')' 8
' *' 353
' sqrt' 17946
'(' 7
'1' 16
'0' 15
')' 8
' =' 284
' ' 220
'5' 20
'0' 15
'*' 9
'sqrt' 26888
'(' 7
'1' 16
'0' 15
').' 568
' Yep' 84194
',' 11
' same' 1852
' thing' 3166
'.' 13
' \n\n' 4710
'So' 4416
',' 11
' after' 1283
' all' 678
' these' 1493
' methods' 5413
' and' 323
' checks' 12341
',' 11
' I' 358
"'m" 2776
' confident' 16506
' that' 429
' the' 279
' simplified' 43799
' form' 1352
' of' 315
' �' 11995
'�' 248
'2' 17
'5' 20
'0' 15
'0' 15
'0' 15
' is' 374
' ' 220
'5' 20
'0' 15
'√' 144336
'1' 16
'0' 15
'.\n\n' 382
'<' 27
'|' 91
'end' 408
'_of' 3575
'_ex' 2702
'planation' 35890
'|' 91
'>' 29
'<|im_end|>' 151645
'\n' 198
テンプレート形式
テキスト全体は以下のテンプレートに従っています。
<|im_start|>system\n
SYSTEM PROMPT
<|im_end|>\n<|im_start|>user\n
QUESTION
<|im_end|>\n<|im_start|>assistant\n<|begin_of_solution|>\n\n
SOLUTION
<|end_of_solution|>\n\n<|begin_of_explanation|>\n\n
EXPLANATION
<|end_of_explanation|><|im_end|>\n
上記のラベルからわかるように、<|begin_of_solution|>\n\n 以前のトークンのラベルはすべて-100であり、損失は計算されません。モデルは、これらのトークンが与えられた場合に何を生成するのが良いかを学習します。つまり、QUESTIONとSOLUTIONが与えられ、<|begin_of_solution|>\n\n のような特殊なトークンが明示された後、どのようなEXPLANATIONを生成するのが良いかを学習します。また、explanationを生成した後、<|end_of_explanation|><|im_end|>\n を出力して対話を終了することも学習します。
RLTのリポジトリでは、<|begin_of_solution|>, <|end_of_solution|>, <|begin_of_explanation|>, <|end_of_explanation|> は特殊トークン（special token）として設定されていなかったため、上記では別々のトークンに分割されています。しかし、モデルはこれらの連続したトークンにどのように対処するかを学習できるはずです。
モデルの使用方法
このモデルをどのように使用するのでしょうか？
# Adapted from https://huggingface.co/Qwen/Qwen2.5-7B-Instruct/blob/main/README.md#quickstart
SYSTEM_PROMPT = "Your role as an assistant involves providing precise and accurate solutions before providing detailed explanations with your full work showing your systematic thinking process leading to each solution. Your explanations should show how you engaged in a comprehensive cycle of analysis, summarizing, exploration, reassessment, reflection, backtracing, and iteration to develop well-considered thinking process. Please structure your response into two main sections: Solution and Explanation. In the Solution section, present your well-thought solution that accurately answers the question. The solution should remain a logical, accurate, concise expression style and detail necessary step needed to reach the conclusion, formatted as follows: <|begin_of_solution|> {final formatted, precise, and clear solution} <|end_of_solution|>. In the Explanation section, comprehensively detail your reasoning process using the specified format: <|begin_of_explanation|> {explanation with steps separated with '\\n\\n'} <|end_of_explanation|> Each step should show detailed considerations leading to your solutions such as analisying questions, summarizing relevant findings, brainstorming new ideas, verifying the accuracy of the current steps, refining any errors, and revisiting previous steps. "
from transformers import AutoModelForCausalLM, AutoTokenizer
model_name = "Team-Promptia/Qwen2.5-7B-RLT-SFT-Stratos-warmup"
model = AutoModelForCausalLM.from_pretrained(
model_name,
torch_dtype="auto",
device_map="auto"
)
tokenizer = AutoTokenizer.from_pretrained(model_name)
QUESTION = "Is the infinite series 1 + 1/2 + 1/3 + 1/4 + ... a convergent infinite series?"
messages = [
{"role": "system", "content": SYSTEM_PROMPT},
{"role": "user", "content": QUESTION}
]
text = tokenizer.apply_chat_template(
messages,
tokenize=False,
add_generation_prompt=True
)
SOLUTION = "No, it is a divergent series."
text += "<|begin_of_solution|>\n\n" + SOLUTION + "\n\n<|end_of_solution|>\n\n<|begin_of_explanation|>\n\n"
model_inputs = tokenizer([text], return_tensors="pt").to(model.device)
generated_ids = model.generate(
**model_inputs,
max_new_tokens=32768
)
generated_ids = [
output_ids[len(input_ids):] for input_ids, output_ids in zip(model_inputs.input_ids, generated_ids)
]
response = tokenizer.batch_decode(generated_ids, skip_special_tokens=True)[0]
システムプロンプトは、このデータセットで規定されているものに修正する必要があります。そして、テストでは、モデルに正解を追加で提供し、特殊トークンで囲み、<|begin_of_explanation|>\n\n を与えて、その後のテキスト、つまりexplanationを生成させる必要があります。
そこで、私たちがモデルに入力したtextは、このようになります。
"<|im_start|>system\nYour role as an assistant involves providing precise and accurate solutions before providing detailed explanations with your full work showing your systematic thinking process leading to each solution. Your explanations should show how you engaged in a comprehensive cycle of analysis, summarizing, exploration, reassessment, reflection, backtracing, and iteration to develop well-considered thinking process. Please structure your response into two main sections: Solution and Explanation. In the Solution section, present your well-thought solution that accurately answers the question. The solution should remain a logical, accurate, concise expression style and detail necessary step needed to reach the conclusion, formatted as follows: <|begin_of_solution|> {final formatted, precise, and clear solution} <|end_of_solution|>. In the Explanation section, comprehensively detail your reasoning process using the specified format: <|begin_of_explanation|> {explanation with steps separated with '\\n\\n'} <|end_of_explanation|> Each step should show detailed considerations leading to your solutions such as analisying questions, summarizing relevant findings, brainstorming new ideas, verifying the accuracy of the current steps, refining any errors, and revisiting previous steps. <|im_end|>\n<|im_start|>user\nIs the infinite series 1 + 1/2 + 1/3 + 1/4 + ... a convergent infinite series?<|im_end|>\n<|im_start|>assistant\n<|begin_of_solution|>\n\nNo, it is a divergent series.\n\n<|end_of_solution|>\n\n<|begin_of_explanation|>\n\n"
あるテストでは、次のresponseが得られました。
"Okay, so I'm trying to figure out if the infinite series 1 + 1/2 + 1/3 + 1/4 + ... is convergent or not. Hmm, I remember that infinite series can either converge to a finite sum or diverge to infinity. But I'm not entirely sure how to determine which category this series falls into.\n\nLet me start by recalling some basic concepts. An infinite series is the sum of the terms of an infinite sequence. In this case, the series is the sum of the reciprocals of the positive integers. So, each term is 1/n where n starts at 1 and goes to infinity. \n\nI think there's a name for this kind of series. It's called the harmonic series. Yeah, the harmonic series. History tells me that this series has been a subject of much study. I remember reading that Euler proved that the harmonic series diverges, but I'm not exactly sure how that proof went.\n\nMaybe I should try to understand what happens when we add up more and more terms. Let's see, the first term is 1. The sum after the first two terms is 1 + 1/2 = 1.5. After three terms, it's 1 + 1/2 + 1/3 ≈ 1.5 + 0.333 ≈ 1.833. After four terms, adding 1/4 (0.25) gives about 2.083. Hmm, so the partial sums are increasing, but do they approach a finite limit or go to infinity?\n\nI think the key here is to look at the behavior of the partial sums as n approaches infinity. If the partial sums stay bounded, the series converges; otherwise, it diverges. In this case, the partial sums keep getting larger and larger. Let me check some more terms. After five terms, adding 1/5 (0.2) gives approximately 2.283. After ten terms, adding up to 1/10 (0.1) would get us around 2.307. Wait, so even after adding many terms, the sum doesn't seem to settle down to a specific value. Instead, it continues to increase, albeit slowly.\n\nI've heard about the concept of the harmonic series' divergence in calculus classes. Maybe there's a test for divergence that can help here. The nth term test states that if the limit of the nth term as n approaches infinity is not zero, then the series diverges. Let's check that. The nth term here is 1/n. The limit as n approaches infinity of 1/n is 0. Wait, so does that mean the series converges? But that contradicts my earlier thought that the partial sums are increasing without bound.\n\nHold on, maybe I mixed up the nth term test. Let me recall: if the limit of the terms is not zero, the series must diverge. If the limit is zero, the test is inconclusive. So in this case, since the limit is zero, the nth term test doesn't tell us anything. So that test is out.\n\nI need another way to determine convergence. Perhaps the integral test? The integral test says that if f(x) is a positive, continuous, and decreasing function for x ≥ 1 and a_n = f(n), then the series ∑a_n converges if and only if the improper integral ∫from 1 to infinity of f(x) dx converges.\n\nLet me consider f(x) = 1/x. The integral of 1/x from 1 to t is ln(t). As t approaches infinity, ln(t) also approaches infinity. Therefore, the integral diverges. By the integral test, this implies that the harmonic series also diverges. That seems to align with what I thought earlier.\n\nBut wait, maybe there's a different approach. Let me think about comparing the harmonic series to another series. For example, grouping terms. Let's see, 1 + (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ... and so on. Each group after the first has four terms, but maybe I can compare these groupings to a geometric series.\n\nWait, the terms 1/3 + 1/4 are both less than 1/2, right? But that might not help. Let me try a different grouping. If I take the first term as 1, then the next two terms (1/2 + 1/3) are each less than 1/2, so their sum is less than 1. Then the next four terms (1/5 to 1/8) are each less than 1/4, so their sum is less than 4*(1/4) = 1. Continuing this way, each subsequent block of 2^k terms contributes less than 1. So the sum of all these blocks would be 1 + 1 + 1 + ... which is a divergent series. But wait, this seems contradictory because if each block contributes 1, then the total sum would be infinite. Therefore, the harmonic series diverges.\n\nAlternatively, maybe using the comparison test. If I compare each term 1/n with 1/2^n. For n ≥ 1, 1/n is greater than 1/2^n for n ≤ 4, but after that, 1/n becomes smaller than 1/2^n. However, since the sum of 1/2^n from n=1 to infinity converges (it's 1), but since 1/n is larger than 1/2^n for the first few terms, and the sum of 1/2^n converges, but actually, wait, the comparison test says that if 0 ≤ a_n ≤ b_n and ∑b_n converges, then ∑a_n also converges. But here, since 1/n ≥ 1/2^n for n=1 to 4, and ∑1/2^n converges, but ∑1/n diverges, so this doesn't help.\n\nWait, maybe I should flip it. Since 1/n is greater than 1/2^n for the first few terms, but 1/2^n converges. However, the problem is that even though 1/2^n converges, the partial sums of 1/n are growing, so the comparison test isn't helpful here.\n\nAnother idea: using the limit comparison test. Let's compare the harmonic series to another series with terms b_n = 1/n. Then compute the limit as n approaches infinity of (a_n / b_n) where a_n is the same as b_n. The limit is 1, and since this limit is a positive finite number, the series ∑a_n and ∑b_n either both converge or both diverge. But since ∑b_n = ∑1/n diverges, then ∑a_n must also diverge. But wait, in this case, a_n = 1/n, so we're comparing it to itself. That seems trivial. Maybe this isn't helpful.\n\nI need to think of another series to compare. Let's consider that starting from n=1, the terms are 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8,... Suppose I group them as 1, (1/2), (1/3 +1/4), (1/5+1/6+1/7+1/8),... etc. Each group after the first has two more terms. Let's approximate each group's sum. The second group (1/3 +1/4) is approximately 0.333 + 0.25 = 0.583, which is less than 0.5. The third group (1/5 +1/6 +1/7 +1/8) is approximately 0.2 + 0.1667 + 0.1429 + 0.125 ≈ 0.6347, which is still less than 1. The fourth group would have 8 terms, each 1/9 to 1/16. The sum of these terms is approximately the integral from 8 to infinity of 1/x dx, which is ln(infinity) - ln(8) → infinity. Wait, but that's not right. Actually, the sum from n=9 to n=16 of 1/n is approximately ln(16) - ln(8) = ln(2) ≈ 0.693. Hmm, so each group after the first contributes more than the previous group. Wait, but my initial approximation was wrong.\n\nLet me correct this. The sum of the third group (1/5 +1/6 +1/7 +1/8) is approximately 0.2 + 0.1667 + 0.1429 + 0.125 ≈ 0.6346. The fourth group (1/9 to 1/16) can be approximated by the integral from 9 to 16 of 1/x dx, which is ln(16) - ln(9) ≈ 2.7726 - 2.1972 ≈ 0.5754. Then the fifth group (1/17 to 1/64) would be ln(64) - ln(16) ≈ 4.1589 - 2.7726 ≈ 1.3863. Ah, so each subsequent group contributes more than the previous one. So the total sum would be 1 + 0.5 + 0.6346 + 0.5754 + 1.3863 + ... which is clearly diverging. Therefore, the harmonic series diverges.\n\nAlternatively, maybe using the Cauchy condensation test. The test states that for a non-negative, non-increasing sequence a_n, the series ∑a_n converges if and only if the condensed series ∑2^n a_{2^n} converges. Let's apply this. For the harmonic series, a_n = 1/n. Then the condensed series would be ∑2^n * (1/(2^n)) = ∑1 = ∑1 from n=1 to infinity, which clearly diverges. Therefore, by the Cauchy condensation test, the original harmonic series must also diverge. \n\nThis seems consistent with all the previous arguments. So, despite the terms approaching zero, the harmonic series doesn't converge; it diverges. Therefore, the answer to the question is no, it's not a convergent infinite series.\n\n<|end_of_explanation|>"
上記の例では、モデルに「無限級数 1 + 1/2 + 1/3 + 1/4 + ... は収束する無限級数ですか？ (Is the infinite series 1 + 1/2 + 1/3 + 1/4 + ... a convergent infinite series?)」と質問し、明確な答え「いいえ、それは発散する級数です (No, it is a divergent series.)」を与えました。
モデルが生成した説明（explanation）を日本語に翻訳すると、次のようになります。
よし、それでは無限級数 1 + 1/2 + 1/3 + 1/4 + … が収束するかどうかを考えてみよう。うーん、無限級数は有限の和に収束するか、無限大に発散するかのどちらかだということは覚えている。でも、この級数がどちらに当てはまるのかはっきりしない。
まずは基本的な概念を思い出してみよう。無限級数とは、無限に続く数列の項をすべて足し合わせたものだ。この場合は、正の整数の逆数の和になる。つまり、各項は 1/n で、n は1から無限大まで進んでいく。
確か、この種の級数には名前があったはずだ。そう、調和級数（harmonic series）という。調和級数は歴史的にも多く研究されてきたもので、オイラーがこの級数が発散することを証明したと読んだ記憶があるけど、その証明の詳細は覚えていない。
もっと項を加えていくとどうなるかを調べてみよう。最初の項は1。2項目までの和は 1 + 1/2 = 1.5。3項目までなら 1 + 1/2 + 1/3 ≈ 1.5 + 0.333 ≈ 1.833。4項目を足すと、1/4（0.25）で、およそ 2.083。うーん、部分和は増え続けているようだが、有限の値に近づいているのか、それとも無限に増えていくのか？
ここで重要なのは、n が無限大に近づくときの部分和の挙動を見ることだ。もし部分和が有限の上限にとどまるなら級数は収束し、そうでなければ発散する。この場合、部分和はどんどん大きくなっている。さらに調べてみよう。5項目を加えると 1/5（0.2）で約 2.283。10項目までなら、1/10（0.1）まで足して約 2.307。つまり、かなり多くの項を足しても、和は特定の値に落ち着かず、緩やかに増え続けている。
微積分の授業で、調和級数が発散するという話を聞いたことがある。ここで何か発散判定法（test for divergence）を使えるかもしれない。n項目のテスト（第n項判定法）は、n が無限大に近づくときの項の極限が 0 でない場合、その級数は発散すると言っていた。ではこの場合、n項目は 1/n。n → ∞ のとき 1/n → 0。え、じゃあこの級数は収束するの？ でもそれはさっきの「和が増え続ける」という観察と矛盾している。
あ、テストの意味を間違えてたかも。正しくは、「項の極限が 0 でない場合は発散確定だが、0 だった場合は判断できない」という内容だった。このケースでは、極限が0なので、このテストでは何もわからない。つまりこの方法では無理。
他の方法を考えてみよう。たとえば積分判定法（integral test）とか。積分判定法は、f(x) が x ≥ 1 において正で連続かつ減少関数で、a_n = f(n) であるとき、無限級数 ∑a_n が収束するかどうかは、定積分 ∫[1, ∞] f(x) dx の収束性と一致する、というもの。
f(x) = 1/x を考えよう。この積分 ∫[1, t] 1/x dx = ln(t)。t → ∞ のとき ln(t) → ∞。つまり、この積分は発散する。よって積分判定法により、調和級数も発散する。これはさっきの直感と一致する。
別のアプローチも試してみよう。例えば項のグルーピングによる比較をしてみる。1 + (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + … のようにグループ化していく。最初のグループ以降はそれぞれ 2, 4, 8 … の項数を持つ。
1/3 + 1/4 はそれぞれ 1/2 未満なので、合計も 1 未満。その次のグループ（1/5〜1/8）はすべて 1/4 未満だから合計は 4×1/4 = 1 未満。このように、それぞれのグループは 1 未満を加えているが、無限に続くため合計は 1 + 1 + 1 + … と同様に発散する。つまり、調和級数は発散する。
比較判定法（comparison test）も考えてみよう。各項 1/n を 1/2^n と比較する。n ≥ 1 において、1/n ≥ 1/2^n となるのは n ≤ 4 くらいまでで、それ以降は逆転する。だが ∑1/2^n は収束する（幾何級数の和は1）。比較判定法によれば、0 ≤ a_n ≤ b_n かつ ∑b_n が収束すれば ∑a_n も収束する。しかしここでは逆に a_n ≥ b_n なので、この方向の比較は役に立たない。
じゃあ極限比較判定法（limit comparison test）を試そう。a_n = 1/n, b_n = 1/n として、lim_{n→∞} (a_n / b_n) = 1。この値が 0 より大きく有限なので、a_n の級数と b_n の級数はともに収束またはともに発散する。今回は自分自身と比較しているから当たり前だが、やはり ∑1/n は発散する。
もう一度グルーピングに戻ろう。項を 1, (1/2), (1/3 + 1/4), (1/5 + 1/6 + 1/7 + 1/8), … のようにまとめてみる。2つ目のグループ (1/3 + 1/4) ≈ 0.333 + 0.25 ≈ 0.583、3つ目のグループ (1/5〜1/8) ≈ 0.2 + 0.1667 + 0.1429 + 0.125 ≈ 0.6347、つまりまだ 1 未満。
4つ目のグループ (1/9〜1/16) の和は ln(16) - ln(9) ≈ 2.7726 - 2.1972 ≈ 0.5754、5つ目のグループ (1/17〜1/32) の和は ln(32) - ln(16) ≈ 3.4657 - 2.7726 ≈ 0.6931。こうしてみると、各グループの和が少しずつ増えていることがわかる。だから全体としては 1 + 0.5 + 0.6346 + 0.5754 + 1.3863 + … のようになり、やはり発散していく。
最後に、コーシーの圧縮判定法（Cauchy condensation test）を使ってみよう。a_n が非負かつ単調減少なら、∑a_n の収束性は ∑2^n a_{2^n} の収束性と一致する。ここでは a_n = 1/n。すると 2^n * (1/2^n) = 1 となり、∑1 は明らかに発散。よって元の級数も発散する。
これまでのすべての議論から一貫して言えるのは、項は0に近づくにもかかわらず、調和級数は収束しない、すなわち発散するということだ。したがって、最初の問いの答えは「いいえ」、この無限級数は収束しない。
ご覧のとおり、最後には正しく <|end_of_explanation|> が生成されています。
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