n-1-cos-pi-e-n-n-4-

Question Number 133708 by Dwaipayan Shikari last updated on 23/Feb/21

\underset{n = 1}{\sum^{\infty}} \frac{c o s ((π + e) n)}{n^{4}}

Answered by mindispower last updated on 24/Feb/21

\sum_{n ⩾ 1} \frac{s i n (n x)}{n} = I m \sum_{n ⩾ 1} \frac{e^{i n x}}{n} = I m (- l n (1 - e^{i x}))

- a r c t a n (\frac{- s i n (x)}{1 - c o s (x)}))

= - (a r c t a n (\frac{1 - c o s (x)}{- s i n (x)} - \frac{π}{2}) = \frac{π - x}{2}

\Rightarrow \sum_{n ⩾ 1} \frac{- c o s (n x)}{n^{2}} = \frac{π}{2} x - \frac{x^{2}}{4} + c

\Rightarrow - Σ \frac{c o s (n x)}{n^{2}} = \frac{π x}{2} - \frac{x^{2}}{4} - \frac{π^{2}}{6}

- \sum_{n ⩾ 1} \frac{s i n (n x)}{n^{3}} = \frac{π x^{2}}{4} - \frac{x^{3}}{12} - \frac{π^{2}}{6} x

\Rightarrow \sum_{n ⩾ 1} \frac{c o s (n x)}{n^{4}} = \frac{π x^{3}}{12} - \frac{x^{4}}{48} - \frac{π^{2} x^{2}}{12} + c, x = 0 \Rightarrow

c = ζ (4)

x = π + e

\sum_{n ⩾ 1} \frac{c o s (π + e) n}{n^{4}} = \frac{π}{12} {(π + e)}^{3} - \frac{{(π + e)}^{4}}{48} - \frac{π^{2}}{12} {(π + e)}^{2} + ζ (4)

Commented by Dwaipayan Shikari last updated on 24/Feb/21

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