calculate-n-1-sin-npi-3-2n-1-2-

Question Number 191369 by mnjuly1970 last updated on 23/Apr/23

c a l c u l a t e

ϕ = \underset{n = 1}{\sum^{\infty}} \frac{s i n (\frac{n π}{3})}{{(2 n + 1)}^{2}} = ?

Answered by namphamduc last updated on 24/Apr/23

S = \underset{n = 1}{\sum^{\infty}} \frac{\sin (\frac{n π}{3})}{{(2 n + 1)}^{2}} = ℑ \underset{n = 1}{\sum^{\infty}} \frac{{(e^{\frac{i π}{3}})}^{n}}{{(2 n + 1)}^{2}}

\tanh^{- 1} (x) = \underset{n = 0}{\sum^{\infty}} \frac{x^{2 n + 1}}{2 n + 1} \Rightarrow \frac{\tanh^{- 1} (x)}{x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{2 n}}{2 n + 1} \Rightarrow \frac{\tanh^{- 1} (x)}{x} - 1 = \underset{n = 1}{\sum^{\infty}} \frac{x^{2 n}}{2 n + 1}

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{x^{2 n}}{{(2 n + 1)}^{2}} = \frac{1}{2 x} ({Li}_{2} (x) - {Li}_{2} (- x)) - 1

\Rightarrow S = ℑ (\frac{1}{2 e^{\frac{i π}{6}}} ({Li}_{2} (e^{\frac{i π}{6}}) - {Li}_{2} (- e^{\frac{i π}{6}})))

= ℑ (\frac{1}{(\sqrt{3} + i)} (\frac{π^{2}}{6} + i \frac{4}{3} G)) = ℑ (\frac{1}{4} (\sqrt{3} - i) (\frac{π^{2}}{6} + i \frac{4}{3} G))

= - \frac{π^{2}}{24} + \frac{G}{\sqrt{3}}

Commented by mnjuly1970 last updated on 24/Apr/23