nice-calculus-prove-that-n-0-2-1-n-2n-1-2-pi-2-8-1-2-ln-2-1-2-

Question Number 135944 by mnjuly1970 last updated on 17/Mar/21

\dots . . n i c e c a l c u l u s \dots

p r o v e t h a t :

\underset{n = 0}{\sum^{\infty}} \frac{{(\sqrt{2} - 1)}^{n}}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8} - \frac{1}{2} {l n}^{2} (1 + \sqrt{2}) \dots

\dots \dots \dots \dots \dots . ✓

Answered by mindispower last updated on 19/Mar/21

h e l l o s i r s u r f o r t h i s r e s u l t ?

w e c a n

\sum_{n ⩾ 0} \frac{{(\sqrt{2} - 1)}^{n}}{{(2 n + 1)}^{2}} = \frac{1}{2} \sum_{n ⩾ 0} 2 \frac{{(\sqrt{\sqrt{2} - 1})}^{2 n}}{{(2 n + 1)}^{2}}

= \frac{1}{2 . \sqrt{\sqrt{2} - 1}} \sum_{n ⩾ 0} \frac{{(\sqrt{\sqrt{2} - 1})}^{2 n + 1}}{{(2 n + 1)}^{2}}

= \frac{1}{2 \sqrt{\sqrt{2} - 1}} (\sum_{n ⩾ 1} \frac{{(\sqrt{\sqrt{2} - 1})}^{n}}{n^{2}} - \sum_{n ⩾ 1} \frac{{(- \sqrt{\sqrt{2} - 1})}^{n}}{n^{2}})

= \frac{1}{2 \sqrt{\sqrt{2} - 1}} ({L i}_{2} (\sqrt{\sqrt{2} - 1}) - {L i}_{2} (- \sqrt{\sqrt{2} - 1})