prove-n-1-1-n-1-H-2n-2n-1-pi-8-ln-2-

Question Number 134016 by mnjuly1970 last updated on 26/Feb/21

? p r o v e : \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} H_{2 n}}{2 n + 1} = \frac{π}{8} l n (2) . .

Answered by mnjuly1970 last updated on 27/Feb/21

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

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

= - \frac{1}{2} \int_{0}^{1} \frac{l n (1 - i x)}{1 - i x} d x - \frac{1}{2} \int_{0}^{1} \frac{l n (1 + i x)}{1 + i x} d x

= \frac{1}{4 i} {[{l n}^{2} (1 - x i)]}_{0}^{1} - \frac{1}{4 i} {[{l n}^{2} (1 + x i)]}_{0}^{1}

= \frac{1}{4 i} {(l n (\sqrt{2} e^{\frac{- π i}{4}}))}^{2} - \frac{1}{4 i} {(l n (\sqrt{2} e^{\frac{π i}{4}}))}^{2}

= \frac{1}{4 i} [{(l n (\sqrt{2}) - \frac{i π}{4})}^{2} - {(l n \sqrt{2)} + \frac{i π}{4})}^{2}]

= \frac{1}{4 i} (- 4 l n (\sqrt{2}) (\frac{i π}{4})) = \frac{- π}{8} l n (2)

∴ S = \frac{π}{8} l n (2) \dots ✓ ✓

\dots m . n \dots