"Show that :∑m=1∞∑n=2m−12m−1m(2n+1)(2n+2)=1−γ" was answered by students like you

leciazyg

Answered question

2022-02-25

Show that :

\sum_{m = 1}^{\infty} \sum_{n = 2^{m - 1}}^{2^{m} - 1} \frac{m}{(2 n + 1) (2 n + 2)} = 1 - γ

Elsie Tillman

Beginner2022-02-26Added 5 answers

\sum_{m = 1}^{\infty} \sum_{n = 2^{m} - 1} m (\frac{1}{2 n + 1} - \frac{1}{2 n + 2})

lim_{N \to \infty} \sum_{m = 1}^{N} (\sum_{n = 2^{m}}^{2^{m + 1} - 1} \frac{m}{n + 1} - \sum_{n = 2^{m - 1}}^{2^{m} - 1} \frac{m}{n + 1})

= lim_{N \to \infty} \sum_{m = 1}^{n} (\sum_{n = 2^{m}}^{2^{m + 1} - 1} \frac{m}{n + 1} - \sum_{n = 2^{m - 1}}^{2^{m} - 1} \frac{m}{n + 1})

= lim_{N \to \infty} (\sum_{n = 2^{N}}^{2^{N + 1} - 1} \frac{N + 1}{n + 1} - \frac{1}{2} - \sum_{n = 2}^{2^{N + 1} - 1} \frac{1}{n + 1})

= lim_{N \to \infty} ((N + 1) (\log (2) + O (2^{- N})) + 1 - \sum_{n = 1}^{2^{N + 1}} \frac{1}{n})

= 1 - γ

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