Sum over k from 1 to Infinity of Zeta of 2k Over Odd Powers of 2/Proof 3

Theorem

$\ds \sum_{k \mathop = 1}^\infty \dfrac {\map \zeta {2k} } {2^{2k - 1} }$

$=$

$\ds \dfrac {\map \zeta {2 } } 2 + \dfrac {\map \zeta {4 } } {2^3} + \dfrac {\map \zeta {6 } } {2^5} + \dfrac {\map \zeta {8 } } {2^7} + \cdots$

$\ds $

$=$

$\ds 1$

$\ds \pi \cot \pi z = \dfrac 1 z - 2 \sum_{k \mathop = 1}^\infty \map \zeta {2 k} z^{2 k - 1}$

Setting $z = \dfrac 1 2$:

$\ds \pi \map \cot {\dfrac \pi 2}$

$=$

$\ds 2 - 2 \sum_{k \mathop = 1}^\infty \dfrac {\map \zeta {2k} } {2^{2k - 1} }$

$\ds \leadsto \ \ $

$\ds 0$

$=$

$\ds 2 - 2 \sum_{k \mathop = 1}^\infty \dfrac {\map \zeta {2k} } {2^{2k - 1} }$

$\ds \leadsto \ \ $

$\ds 2 \sum_{k \mathop = 1}^\infty \dfrac {\map \zeta {2k} } {2^{2k - 1} }$

$=$

$\ds 2$

$\ds $

$=$

$\ds 1$

dividing both sides by $2$

Hence:

$\ds \sum_{k \mathop = 1}^\infty \dfrac {\map \zeta {2k} } {2^{2k - 1} }$

$=$

$\ds 1$

$\blacksquare$