nice-calculus-find-the-value-of-n-1-1-n-sin-2-n-n-

Question Number 138683 by mnjuly1970 last updated on 16/Apr/21

\dots n i c e . . \dots \dots c a l c u l u s \dots

f i n d t h e v a l u e o f :

Θ = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} {s i n}^{2} (n)}{n} = ?

\dots \dots \dots \dots \dots \dots \dots \dots .

Answered by Dwaipayan Shikari last updated on 16/Apr/21

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} {s i n}^{2} n}{n} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n} - \frac{{(- 1)}^{n} c o s (2 n)}{2 n}

= - \frac{l o g (2)}{2} + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} c o s (2 n)}{n}

= - \frac{l o g (2)}{2} + \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} e^{2 i n}}{n} + \frac{{(- 1)}^{n + 1} e^{- 2 i n}}{n}

= - \frac{l o g (2)}{2} + \frac{1}{4} (l o g (1 + e^{2 i}) + l o g (1 + e^{- 2 i}))

= - \frac{l o g (2)}{2} + \frac{1}{4} l o g (2 + 2 c o s (2)) = - \frac{l o g (2)}{2} + \frac{1}{2} l o g (s i n 1)

= \frac{l o g (\frac{s i n (1)}{2})}{2}