雑談はここに書け!【67】 (481レス)
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: 09/19(金)07:36
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321: [] 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
追加 何故ポールエルデシュは収束する級数 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 日本語の倍らしい 英語検索 による概要
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