Question-159233

Question Number 159233 by mnjuly1970 last updated on 14/Nov/21

Answered by mindispower last updated on 16/Nov/21

A_{2 n} = \sum_{k ⩾ 1} \frac{1}{{(2 k)}^{2 n}} = \frac{1}{2^{2 n}} . ζ (2 n)

\frac{\frac{ζ (2 n)}{2^{2 n}}}{2 n (2 n + 1)} = - \frac{ζ (2 n)}{2^{2 n} (2 n + 1)} + \frac{ζ (2 n)}{2^{2 n} . (2 n)}

\sum_{k ⩾ 1} ζ (2 k) x^{2 k} = - \frac{π x}{2} c o t (π x) + \frac{1}{2}

\sum_{k ⩾ 1} \frac{ζ (2 k) x^{2 k}}{2 k} = \int_{0}^{x} (- \frac{π}{2} c o t (π x) + \frac{1}{2 x}) d x

= \lim_{t \to 0} \int_{t}^{x} (- \frac{π}{2} c o t (π x) + \frac{1}{2 x}) d x =

= \lim_{t \to 0} \int_{t}^{x} (- \frac{1}{2} l n (s i n (π x)) + \frac{1}{2} l n (x))

\frac{1}{2} (l n (\frac{x}{s i n π x}) - \lim_{t \to 0} l n (\frac{t}{s i n (π t)})

= l n (\sqrt{\frac{x}{s i n (π x)}}) + l n (\sqrt{π})

\sum_{k ⩾ 1} \frac{ζ (2 n)}{2^{2 n} (2 n)} = l n (\sqrt{\frac{1}{2}}) + l n (\sqrt{π})

\sum_{n ⩾ 1} \frac{ζ (2 n)}{2^{2 n} (2 n + 1)} = 2 \int_{0}^{\frac{1}{2}} \sum_{n ⩾ 1} ζ (2 n) x^{2 n} d x

= 2 \int_{0}^{\frac{1}{2}} (\frac{1}{2} - \frac{π}{2} x c o t (π x)) d x = \frac{1}{2} - \frac{1}{π} \int_{0}^{\frac{π}{2}} u c o t (u) d u

= (\frac{1}{2} + \frac{1}{π} \int_{0}^{\frac{π}{2}} l n (s i n (u)) d u) = \frac{1}{2} + \frac{1}{π} . - \frac{π}{2} l n (2) = \frac{1}{2} (1 - l n (2))

S = \frac{1}{2} (l n (π) - l n (2)) - \frac{1}{2} (1 - l n (2))

= \frac{1}{2} (l n (π) - 1)

Commented by mnjuly1970 last updated on 15/Nov/21

v e r y n i c e s i r p o w e r \dots t h x a l o t

Commented by mindispower last updated on 15/Nov/21

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