1-find-S-x-n-1-cos-nx-n-2-find-n-1-1-n-n-

Question Number 38109 by maxmathsup by imad last updated on 21/Jun/18

1) f i n d S (x) = \sum_{n = 1}^{\infty} \frac{c o s (n x)}{n}

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

Commented by abdo.msup.com last updated on 30/Jun/18

w e h a v e S (x) = R e (\sum_{n = 1}^{\infty} \frac{e^{i n x}}{n}) = R e (w (x))

w (x) = \sum_{n = 1}^{\infty} \frac{e^{i n x}}{n} \Rightarrow w^{^{'}} (x) = i \sum_{n = 1}^{\infty} {(e^{i x})}^{n}

= i e^{i x} \sum_{n = 1}^{\infty} {(e^{i x})}^{n - 1} = {i e}^{i x} \sum_{n = 0}^{\infty} {(e^{i x})}^{n}

= i \frac{e^{i x}}{1 - e^{i x}} = i \frac{e^{i x}}{1 - c o s x - i s i n x}

= i \frac{e^{i x}}{2 {s i n}^{2} (\frac{x}{2}) - 2 i s i n (\frac{x}{2}) c o s (\frac{x}{2})}

= \frac{i}{- 2 i s i n (\frac{x}{2})} \frac{e^{i x}}{e^{i \frac{x}{2}}}

= \frac{- 1}{2 s i n (\frac{x}{2})} e^{i \frac{x}{2}}

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

= - \frac{1}{2} c o t a n (\frac{x}{2}) - \frac{i}{2} \Rightarrow

w (x) = - \frac{1}{2} \int \frac{c o s (\frac{x}{2})}{s i n (\frac{x}{2})} - \frac{i x}{2} + c

= - l n ∣ s i n (\frac{x}{2}) ∣ - \frac{i x}{2} + c

w (π) = - \frac{i π}{2} + c = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2)

\Rightarrow c = \frac{i π}{2} - l n (2) \Rightarrow

w (x) = - l n ∣ s i n (\frac{x}{2}) ∣ - \frac{i x}{2} + \frac{i π}{2} - l n (2)

S (x) = R e (w (x)) = - l n ∣ s i n (\frac{x}{2}) ∣ - l n (2) .

a l s o w e g e t \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n} = \frac{π - x}{2} .

Commented by abdo.msup.com last updated on 30/Jun/18

2) w e h a v e p r o v e f t h a t

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} = l n (1 + x) i f ∣ x ∣⩽ 1 a n d

x \neq - 1 \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2) \Rightarrow

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2) .