n-1-n-ln-2n-1-2n-1-1-

Question Number 156776 by mnjuly1970 last updated on 15/Oct/21

\underset{n = 1}{\sum^{\infty}} (n l n (\frac{2 n + 1}{2 n - 1}) - 1) = ?

Commented by CAIMAN last updated on 15/Oct/21

1/2-1/2 ln2

Answered by qaz last updated on 15/Oct/21

\underset{n = 1}{\sum^{\infty}} (nln \frac{2 n + 1}{2 n - 1} - 1)

= \underset{n = 1}{\sum^{\infty}} (nln (2 n + 1) - nln (2 n - 1) - 1)

= \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} (\frac{n}{x + 2 n} - \frac{n}{x - 2 n} - 1) dx

= \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} (\frac{n}{x + 2 n} - \frac{n}{x - 2 n} - 1) dx

= \frac{1}{4} \int_{0}^{1} x^{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} - {(\frac{x}{2})}^{2}} dx

= \frac{1}{4} \int_{0}^{1} x^{2} \frac{1 - π (\frac{x}{2}) \cot \frac{π x}{2}}{2 {(\frac{x}{2})}^{2}} dx

= \frac{1}{4} \int_{0}^{1} (2 - π xcot \frac{π x}{2}) dx

= \frac{1}{4} (2 - 2 \ln 2)

= \frac{1}{2} (1 - \ln 2)

Commented by mnjuly1970 last updated on 15/Oct/21

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