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Question Number 98118 by abdomathmax last updated on 11/Jun/20

calculate \sum_{n = 1}^{\infty} (ξ (2 n) - 1) x^{2 n}

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

Answered by maths mind last updated on 11/Jun/20

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

= \sum_{n ⩾ 1} . \sum_{k ⩾ 2} \frac{x^{2 n}}{k^{2 n}}

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

= \sum_{k ⩾ 2} (\frac{x^{2}}{k^{2}}) . \frac{1}{1 - \frac{x^{2}}{k^{2}}}

= \sum_{k ⩾ 2} \frac{x^{2}}{k^{2} - x^{2}} = S (x)

f o r x \notin Z

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

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

Commented by maths mind last updated on 11/Jun/20

x \in] 1, 2 [

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

thank you sir