1) find S(x) = Σ_(n=1) ^∞ ((cos(nx))/n) 2) find Σ


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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
$we have S(x)=Re(Σ_(n=1) ^∞ (e^(inx) /n))=Re(w(x)) w(x)=Σ_(n=1) ^∞ (e^(inx) /n) ⇒w^′ (x)=i Σ_(n=1) ^∞ (e^(ix) )^n =i e^(ix) Σ_(n=1) ^∞ (e^(ix) )^(n−1) =ie^(ix) Σ_(n=0) ^∞ (e^(ix) )^n =i(e^(ix) /(1−e^(ix) )) =i (e^(ix) /(1−cosx −isinx)) =i (e^(ix) /(2sin^2 ((x/2))−2isin((x/2))cos((x/2)))) = (i/(−2isin((x/2)))) (e^(ix) /e^(i(x/2)) ) =((−1)/(2 sin((x/2)))) e^(i(x/2)) = ((−1)/(2sin((x/2)))){ cos((x/2)) +i sin((x/2))} =−(1/2)cotan((x/2)) −(i/2) ⇒ w(x)=−(1/2) ∫ ((cos((x/2)))/(sin((x/2)))) −((ix)/2) +c =−ln∣sin((x/2))∣ −((ix)/2) +c w(π) =−((iπ)/2) +c =Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) ⇒c=((iπ)/2) −ln(2) ⇒ w(x)=−ln∣ sin((x/2))∣−((ix)/2) +((iπ)/2) −ln(2) S(x)=Re(w(x))=−ln∣sin((x/2))∣−ln(2). also we get Σ_(n=1) ^∞ ((sin(nx))/n) =((π−x)/2) .$
$w e h a v e S (x) = R e (\sum_{n = 1}^{\infty} \frac{e^{i n x}}{n}) = R e (w (x))$ $w (x) = \sum_{n = 1}^{\infty} \frac{e^{i n x}}{n} \Rightarrow w^{^{'}} (x) = i \sum_{n = 1}^{\infty} {(e^{i x})}^{n}$ $= i e^{i x} \sum_{n = 1}^{\infty} {(e^{i x})}^{n - 1} = {i e}^{i x} \sum_{n = 0}^{\infty} {(e^{i x})}^{n}$ $= i \frac{e^{i x}}{1 - e^{i x}} = i \frac{e^{i x}}{1 - c o s x - i s i n x}$ $= i \frac{e^{i x}}{2 {s i n}^{2} (\frac{x}{2}) - 2 i s i n (\frac{x}{2}) c o s (\frac{x}{2})}$ $= \frac{i}{- 2 i s i n (\frac{x}{2})} \frac{e^{i x}}{e^{i \frac{x}{2}}}$ $= \frac{- 1}{2 s i n (\frac{x}{2})} e^{i \frac{x}{2}}$ $= \frac{- 1}{2 s i n (\frac{x}{2})} {c o s (\frac{x}{2}) + i s i n (\frac{x}{2})}$ $= - \frac{1}{2} c o t a n (\frac{x}{2}) - \frac{i}{2} \Rightarrow$ $w (x) = - \frac{1}{2} \int \frac{c o s (\frac{x}{2})}{s i n (\frac{x}{2})} - \frac{i x}{2} + c$ $= - l n ∣ s i n (\frac{x}{2}) ∣ - \frac{i x}{2} + c$ $w (π) = - \frac{i π}{2} + c = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2)$ $\Rightarrow c = \frac{i π}{2} - l n (2) \Rightarrow$ $w (x) = - l n ∣ s i n (\frac{x}{2}) ∣ - \frac{i x}{2} + \frac{i π}{2} - l n (2)$ $S (x) = R e (w (x)) = - l n ∣ s i n (\frac{x}{2}) ∣ - l n (2) .$ $a l s o w e g e t \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n} = \frac{π - x}{2} .$
Commented by abdo.msup.com last updated on 30/Jun/18

$2) w e h a v e p r o v e f t h a t$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} = l n (1 + x) i f ∣ x ∣⩽ 1 a n d$ $x \neq - 1 \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2) \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2) .$
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