Σ_(n=1) ^∞ ((cos(n))/n^2 )


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Question Number 132434 by Dwaipayan Shikari last updated on 14/Feb/21

$\underset{n = 1}{\sum^{\infty}} \frac{c o s (n)}{n^{2}}$
	Answered by mnjuly1970 last updated on 14/Feb/21

	$s o l u t i o n :$ $\underset{n = 1}{\sum^{\infty}} \frac{e^{i n} + e^{- i n}}{2 n^{2}} = \frac{1}{2} [{l i}_{2} (e^{i}) + {l i}_{2} (e^{- i})]$ $= \frac{1}{2} ({l i}_{2} (e^{i}) + {l i}_{2} (\frac{1}{e^{i}}))$ $\underset{- ζ (2) - \frac{1}{2} {l n}^{2} (- x)}{\overset{{l i}_{2} (x) + {l i}_{2} (\frac{1}{x})}{=}} \frac{1}{2} (- \frac{π^{2}}{6} - \frac{1}{2} {l n}^{2} (- e^{i}))$ $= \frac{- π^{2}}{12} - \frac{1}{4} {(2 l n (i) + i)}^{2}$ $= \frac{- π^{2}}{12} - \frac{1}{4} (i {(π + 1)}^{2}) = \frac{- π^{2}}{12} + \frac{{(π + 1)}^{2}}{4}$
	Commented by Dwaipayan Shikari last updated on 14/Feb/21

	$G r e a t s i r! t h a n k i n g y o u . .$
	Commented by mnjuly1970 last updated on 14/Feb/21

	$y o u a r e w e l c o m e . . .$
	Answered by mathmax by abdo last updated on 14/Feb/21
	$let ϕ(x)=Σ_(n=1) ^∞ ((sin(nx))/n) we have ∫ ϕ(x)dx=−Σ_(n=1) ^∞ ((cos(nx))/n^2 ) ⇒Σ_(n=1) ^∞ ((cos(nx))/n^2 )=K−∫ ϕ(x)dx ϕ(x)=Im(Σ_(n=1) ^∞ (e^(inx) /n)) but Σ_(n=1) ^∞ (((e^(ix) )^n )/n) =Ψ(e^(ix) )with Ψ(u) =Σ_(n=1) ^∞ (u^n /n) =−ln(1−u) ⇒Ψ(e^(ix) )=−ln(1−e^(ix) ) =−ln(1−cosx−isinx) =−ln(2sin^2 ((x/2))−2isin((x/2))cos((x/2))) =−ln(−2isin((x/2)){cos((x/2))+isin((x/2))}) =−ln(−2)−ln(i)−ln(sin((x/2))−ln(e^((ix)/2) ) =−ln(e^(iπ) )−ln(2)−ln(e^((iπ)/2) )−ln(sin((x/2)))−((ix)/2) =−iπ−((iπ)/2)−((ix)/2) −ln(sin((x/2)))−ln(2) =i(−((3π)/2)−(x/2))−ln(2sin((x/2))) ⇒ϕ(x)=−(x/2)−((3π)/2) ⇒ ∫ ϕ(x)dx =−(x^2 /4)−((3π)/2)x ⇒Σ_(n=1) ^∞ ((cos(nx))/n^2 )=k+(x^2 /4)+((3π)/2)x x=0 ⇒(π^2 /6)=k ⇒Σ_(n=1) ^∞ ((cos(nx))/n^2 ) =(x^2 /4)+((3π)/2)x +(π^2 /6) x=1 ⇒Σ_(n=1) ^∞ ((cos(n))/n^2 ) =(1/4)+((3π)/2)+(π^2 /6)$
	$let φ (x) = \sum_{n = 1}^{\infty} \frac{\sin (nx)}{n} we have \int φ (x) dx = - \sum_{n = 1}^{\infty} \frac{\cos (nx)}{n^{2}}$ $\Rightarrow \sum_{n = 1}^{\infty} \frac{\cos (nx)}{n^{2}} = K - \int φ (x) dx$ $φ (x) = Im (\sum_{n = 1}^{\infty} \frac{e^{inx}}{n}) but \sum_{n = 1}^{\infty} \frac{{(e^{ix})}^{n}}{n} = Ψ (e^{ix}) with$ $Ψ (u) = \sum_{n = 1}^{\infty} \frac{u^{n}}{n} = - \ln (1 - u) \Rightarrow Ψ (e^{ix}) = - \ln (1 - e^{ix})$ $= - \ln (1 - cosx - isinx) = - \ln ({2 \sin}^{2} (\frac{x}{2}) - 2 isin (\frac{x}{2}) \cos (\frac{x}{2}))$ $= - \ln (- 2 isin (\frac{x}{2}) {\cos (\frac{x}{2}) + isin (\frac{x}{2})})$ $= - \ln (- 2) - \ln (i) - \ln (\sin (\frac{x}{2}) - \ln (e^{\frac{ix}{2}})$ $= - \ln (e^{i π}) - \ln (2) - \ln (e^{\frac{i π}{2}}) - \ln (\sin (\frac{x}{2})) - \frac{ix}{2}$ $= - i π - \frac{i π}{2} - \frac{ix}{2} - \ln (\sin (\frac{x}{2})) - \ln (2)$ $= i (- \frac{3 π}{2} - \frac{x}{2}) - \ln (2 \sin (\frac{x}{2})) \Rightarrow φ (x) = - \frac{x}{2} - \frac{3 π}{2} \Rightarrow$ $\int φ (x) dx = - \frac{x^{2}}{4} - \frac{3 π}{2} x \Rightarrow \sum_{n = 1}^{\infty} \frac{\cos (nx)}{n^{2}} = k + \frac{x^{2}}{4} + \frac{3 π}{2} x$ $x = 0 \Rightarrow \frac{π^{2}}{6} = k \Rightarrow \sum_{n = 1}^{\infty} \frac{\cos (nx)}{n^{2}} = \frac{x^{2}}{4} + \frac{3 π}{2} x + \frac{π^{2}}{6}$ $x = 1 \Rightarrow \sum_{n = 1}^{\infty} \frac{\cos (n)}{n^{2}} = \frac{1}{4} + \frac{3 π}{2} + \frac{π^{2}}{6}$
	Commented by Dwaipayan Shikari last updated on 15/Feb/21

	$G r e a t s i r!$
	Commented by mathmax by abdo last updated on 20/Feb/21

	$thanks$
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