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Question Number 133432 by Dwaipayan Shikari last updated on 22/Feb/21

\frac{s i n \sqrt{π}}{1^{3}} + \frac{s i n \sqrt{4 π}}{2^{3}} + \frac{s i n \sqrt{9 π}}{3^{3}} + \frac{s i n \sqrt{16 π}}{4^{3}} + \dots . = \frac{π \sqrt{π}}{12} (1 - 3 \sqrt{π} + 2 π)

Answered by mindispower last updated on 24/Feb/21

\sum_{n ⩾ 0} \frac{s i m (n x)}{n} = \frac{π - x}{2_{}}

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

\Rightarrow \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 = f (x)

\frac{s i n (\sqrt{π})}{1^{3}} + \frac{s i n (\sqrt{4 π})}{2^{2}} + \dots . . + \frac{s i n (\sqrt{n^{2} π})}{n^{2}} + \dots

= Σ \frac{s i n (n \sqrt{π})}{n^{3}} = f (\sqrt{π}) = - \frac{π^{2}}{4} + \frac{π \sqrt{π}}{12} + \frac{π^{2} \sqrt{π}}{6}

= \frac{π \sqrt{π}}{12} (1 - 3 \sqrt{π} + 2 π)