advanced-calculus-n-1-sin-n-n-3-

Question Number 139479 by mnjuly1970 last updated on 27/Apr/21

\dots . a d v a n c e d \dots . ★ ★ ★ \dots . . c a l c u l u s \dots . .

:= \underset{n = 1}{\sum^{\infty}} {(\frac{s i n (n)}{n})}^{3} = ?

Answered by Dwaipayan Shikari last updated on 27/Apr/21

\underset{n = 1}{\sum^{\infty}} \frac{{s i n}^{3} (n)}{n^{3}} = \frac{3}{4} \underset{n = 1}{\sum^{\infty}} \frac{s i n (n)}{n^{3}} - \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{s i n (3 n)}{n^{3}}

= \frac{3}{4} (\frac{1}{12} + \frac{π^{2}}{6} - \frac{π}{4}) - \frac{1}{4} (\frac{3}{12} + \frac{π^{2}}{2} - \frac{9 π}{4})

= \frac{9 π}{16} - \frac{3 π}{16} = \frac{3 π}{8} (\underset{n = 1}{\sum^{\infty}} \frac{s i n (n θ)}{n^{3}} = \frac{θ^{3}}{12} + \frac{π^{2} θ}{6} - \frac{π θ^{2}}{4})

Commented by mnjuly1970 last updated on 27/Apr/21

m e r c e y m r D w a i p a y a n . .