let-0-lt-lt-pi-find-n-1-cos-n-n-2-find-n-1-sin-n-n-

Question Number 32938 by abdo imad last updated on 06/Apr/18

l e t 0 < θ < π f i n d \sum_{n = 1}^{\infty} \frac{c o s (n θ)}{n}

2) f i n d \sum_{n = 1}^{\infty} \frac{s i n (n θ)}{n}

Commented by abdo imad last updated on 09/Apr/18

l e t p u t w (z) = \sum_{n = 0}^{\infty} z^{n} w i t h ∣ z ∣⩽ 1 a n d \in C - R w e h a v e

w (z) = \frac{1}{1 - z} \Rightarrow \int w (z) d z = - l n (1 - z) \Rightarrow

\sum_{n = 0}^{\infty} \frac{z^{n + 1}}{n + 1} = - l n (1 - z) \Rightarrow \sum_{n = 1}^{\infty} \frac{z^{n}}{n} = - l n (1 - z)

a n d f o r z = e^{i θ} w e g e t \sum_{n = 1}^{\infty} \frac{e^{i θ}}{n} = - l n (1 - e^{i θ})

b u t l n (1 - e^{i θ}) = l n (1 - c o s θ - i s i n θ)

= l n (2 {s i n}^{2} (\frac{θ}{2}) - 2 i c o s (\frac{θ}{2}) s i n (\frac{θ}{2}))

= l n (- 2 i s i n (\frac{θ}{2}) e^{i \frac{θ}{2}}) = i \frac{θ}{2} + l n (- 2 i) + l n (s i n (\frac{θ}{2}))

= l n (2) + l n (s i n (\frac{θ}{2})) + i \frac{θ}{2} + l n (e^{- i \frac{π}{2}})

= l n (2 s i n (\frac{θ}{2})) + i \frac{θ - π}{2} \Rightarrow \sum_{n = 1}^{\infty} \frac{e^{i n θ}}{n}

= - l n (2 s i n (\frac{θ}{2})) + i \frac{π - θ}{2} \Rightarrow

\sum_{n = 1}^{\infty} \frac{c o s (n θ)}{n} = - l n (2 s i n (\frac{θ}{2})) a n d

\sum_{n = 1}^{\infty} \frac{s i n (n θ)}{n} = \frac{π - θ}{2} .

Commented by prof Abdo imad last updated on 09/Apr/18

\sum_{n = 1}^{\infty} \frac{e^{i n θ}}{n} = - l n (1 - e^{i θ}) .