1-2-1-4-3-1-9-4-1-16-5-1-25-6-1-36-7-1-49-8-1-64-exp-2-2-pi-2-12-log-2-

Question Number 142035 by Dwaipayan Shikari last updated on 25/May/21

\frac{1}{2^{1 / 4}} . \frac{3^{1 / 9}}{4^{1 / 16}} . \frac{5^{1 / 25}}{6^{1 / 36}} . \frac{7^{1 / 49}}{8^{1 / 64}} \dots = e x p (- \frac{ζ^{'} (2)}{2} - \frac{π^{2}}{12} \log (2))

Answered by mindispower last updated on 25/May/21

= \prod_{k ⩾ 1} \frac{{(1 + 2 k)}^{\frac{1}{{(2 k + 1)}^{2}}}}{{(2 k)}^{\frac{1}{{(2 k)}^{2}}}} = Ψ

l n (Ψ) = \sum_{k ⩾ 1} \frac{l n (2 k + 1)}{(2 k + 1)} - \sum_{k ⩾ 1} \frac{l n (2 k)}{4 k^{2}} = A - B

ζ (x) = \sum_{n ⩾ 1} \frac{1}{n^{x}}, ζ^{'} (x) = - \sum_{n ⩾ 1} \frac{l n (n)}{n^{x}}

B = \sum_{k ⩾ 1} \frac{l n (2)}{4} . \frac{1}{k^{2}} + \frac{1}{4} \sum_{k ⩾ 1} \frac{l n (k)}{k^{2}} = \frac{l n (2)}{4} ζ (2) - \frac{ζ^{'} (2)}{4}

A = \sum_{k ⩾ 1} (\frac{l n (k)}{k^{2}} - \frac{l n (2 k)}{4 k^{2}}) = - ζ^{'} (2) - B

l n (Ψ) = - \frac{1}{2} ζ^{'} (2) - \frac{l n (2)}{2} ζ (2) = - \frac{ζ^{'} (2)}{2} - \frac{π^{2} l n (2)}{12}

Ψ = e x p (\frac{- ζ^{'} (2)}{2} - \frac{π^{2} l n (2)}{12})

Commented by Dwaipayan Shikari last updated on 25/May/21

T h a n k s s i r . G r e a t!

Commented by mindispower last updated on 26/May/21

p l e a s u r s i r