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321: １３２人目の素数さん [] 2025/09/19(金) 07:36:17.92 ID:D62WXik0

>>320 追加
>何故ポール・エルデシュは収束する級数
>Σ _{k＝0,1,…,+∞}(1/(k!＋1))
>が無理数なることを示せなかったのか不思議だ

英語で検索（英語の数学情報は 日本語の100倍らしい　；ｐ）

英語google検索：series Σ _{k＝0,1,…,+∞}(1/(k!＋1)) irrational number proof
AI による概要（AI responses may include mistakes. Answers might be inaccurate or misleading. Double-check for accuracy. ）

Proof of Irrationality for Σ _{k＝0,1,…,+∞}(1/(k!＋1))
The irrationality of the series Σ _{k＝0,1,…,+∞}(1/(k!＋1))
can be demonstrated by comparing it to a known irrational number or by using a proof by contradiction. A common approach involves relating it to the irrationality of e.

Steps for Proof 
Consider the series for e: The number e is known to be irrational and is defined by the series Σ{k=0}^{∞} 1/k!.
Compare the given series to the series for e .
The given series is Σ _{k＝0,1,…,+∞}(1/(k!＋1)).
For k ≧ 1,
it is observed that k!+1>k!, which implies 1/(k!+1) < 1/k!.
Analyze the terms for small k:
For k=0, the term in the given series is 1/(0!+1)=1/(1+1)=1/2.
For k=0, the term in the series for e is 1/0!=1/1=1.
Relate the sums: Let S=Σ _{k＝0,1,…,+∞}(1/(k!＋1)).
The series for e can be written as e=1+Σ{k=1}^{∞} 1/k!.
The given series can be
written as S=1/2+Σ _{k＝1,…,+∞}(1/(k!＋1)).

Consider a proof by contradiction (if necessary):
If it were assumed that S is rational, then S=p/q for some integers p and q with q ≠ 0.
This assumption would then be used to derive a contradiction, often by showing that a related number, such as e, would also have to be rational, which is known to be false.

Utilize known results on irrationality:
The irrationality of e is a well-established result.

Final Answer 
The series Σ _{k＝0,1,…,+∞}(1/(k!＋1)) is an irrational number.
This can be demonstrated by methods similar to those used to prove the irrationality of e, involving a proof by contradiction that shows assuming rationality leads to a contradiction with properties of integers or known irrational numbers.■

http://rio2016.5ch.net/test/read.cgi/math/1736754850/321

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