advanced-calculus-evaluate-n-2-2n-2-n-

Question Number 125458 by mnjuly1970 last updated on 11/Dec/20

\dots a d v a n c e d c a l c u l u s \dots

e v a l u a t e :::

\underset{n = 2}{\sum^{\infty}} {\frac{ζ (2 n)}{2^{n}}} = ? ?

Answered by Dwaipayan Shikari last updated on 11/Dec/20

\underset{n = 2}{\sum^{\infty}} \underset{k = 1}{\sum^{\infty}} \frac{1}{2^{n} k^{2 n}} = \underset{k = 1}{\sum^{\infty}} \underset{n ⩾ 1}{\sum^{\infty}} \frac{1}{2^{n} {(k^{2})}^{n}} = \underset{k = 1}{\sum^{\infty}} \frac{\frac{1}{4 k^{4}}}{1 - \frac{1}{2 k^{2}}}

= \underset{k = 1}{\sum^{\infty}} \frac{1}{2 k^{2} - 1} = \underset{k = 1}{\sum^{\infty}} \frac{1}{2 k^{2} (2 k^{2} - 1)}

= \sum^{\infty} \frac{1}{2 k^{2} - 1} - \frac{1}{2 k^{2}} = \frac{1}{2} \sum^{\infty} \frac{1}{k^{2} - \frac{1}{2}} - \frac{π^{2}}{12}

= \frac{1}{2 \sqrt{2}} \sum^{\infty} \frac{1}{k - \frac{1}{\sqrt{2}}} - \frac{1}{k + \frac{1}{\sqrt{2}}} - \frac{π^{2}}{12} = \frac{1}{2 \sqrt{2}} (ψ (1 + \frac{1}{\sqrt{2}}) - ψ (1 - \frac{1}{\sqrt{2}})) - \frac{π^{2}}{12}

\sim 0.52

Commented by Dwaipayan Shikari last updated on 11/Dec/20

ζ (2 n) = \frac{{(- 1)}^{n + 1} β_{2 n} {(2 π)}^{2 n}}{2 (2 n)!}

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

β_{n} - B e r n o u l l i N u m b e r

Commented by mnjuly1970 last updated on 11/Dec/20

p e a c e b e u p o n y o u \dots

Commented by Dwaipayan Shikari last updated on 11/Dec/20

I s i t r i g h t ? (M y a n s w e r)

Commented by mnjuly1970 last updated on 11/Dec/20

y e s o f c o u r s e .

b u t o r i g i n a l q u e s t i o n w a s a s b e l o w .

\underset{n = 2}{\sum^{\infty}} \frac{ζ (n)}{2^{n}} \overset{? ?}{=} l o g (2) \dots

Commented by Dwaipayan Shikari last updated on 11/Dec/20

\sum^{\infty} \frac{ζ (n)}{2^{n}} = \underset{k = 1}{\sum^{\infty}} \underset{n = 2}{\sum^{\infty}} \frac{1}{{(2 k)}^{n}} = \underset{k = 1}{\sum^{\infty}} \frac{1}{2 k (2 k - 1)} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + . . = l o g (2)

Commented by mnjuly1970 last updated on 11/Dec/20

p e r f e c t . .

Commented by Dwaipayan Shikari last updated on 11/Dec/20

W i t h p l e a s u r e

Answered by mathmax by abdo last updated on 11/Dec/20

\sum_{n = 2}^{\infty} \frac{ξ (2 n)}{2^{n}} = \sum_{n = 2}^{\infty} \frac{1}{2^{n}} \sum_{k = 1}^{\infty} \frac{1}{k^{2 n}} = \sum_{k = 1}^{\infty} \sum_{n = 2}^{\infty} \frac{1}{2^{n}} \frac{1}{{(k^{2})}^{n}}

= \sum_{k = 1}^{\infty} \sum_{n = 2}^{\infty} {(\frac{1}{{2 k}^{2}})}^{n} = \sum_{k = 1}^{\infty} \sum_{p = 0}^{\infty} {(\frac{1}{{2 k}^{2}})}^{p + 2}

= \sum_{k = 1}^{\infty} \frac{1}{{4 k}^{4}} \sum_{p = 0}^{\infty} {(\frac{1}{{2 k}^{2}})}^{p} = \sum_{k = 1}^{\infty} \frac{1}{{4 k}^{4}} \times \frac{1}{1 - \frac{1}{{2 k}^{2}}}

= \sum_{k = 1}^{\infty} \frac{{2 k}^{2}}{{4 k}^{4} ({2 k}^{2} - 1)} = \sum_{k = 1}^{\infty} \frac{1}{{2 k}^{2} ({2 k}^{2} - 1)}

= \sum_{k = 1}^{\infty} (\frac{1}{{2 k}^{2} - 1} - \frac{1}{{2 k}^{2}}) = \sum_{k = 1}^{\infty} \frac{1}{{2 k}^{2} - 1} - \frac{1}{2} \times \frac{π^{2}}{6}

= \sum_{k = 1}^{\infty} \frac{1}{{2 k}^{2} - 1} - \frac{π^{2}}{12} rest to find value of \sum_{k = 1}^{\infty} \frac{1}{{2 k}^{2} - 1}

\dots be continued \dots

Commented by mnjuly1970 last updated on 11/Dec/20

t h a n k s a l o t s i r \dots

Commented by mathmax by abdo last updated on 11/Dec/20

you are welcome sir .