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Question Number 109639 by Ar Brandon last updated on 24/Aug/20

	Answered by Dwaipayan Shikari last updated on 25/Aug/20

	$S = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} = (- 1) \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n^{2}} = (- 1) (\frac{1}{1} - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . . .)$ $S_{n} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . .$ $S_{n} - S = 2 (\frac{1}{2^{2}} + \frac{1}{4^{2}} + . . . .) = 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{2}}$ $\sum^{\infty} \frac{1}{n^{2}} - \sum^{\infty} \frac{1}{2 n^{2}} = S$ $\frac{1}{2} \sum^{\infty} \frac{1}{n^{2}} = S$ $\frac{1}{2} . \frac{π^{2}}{6} = S$ $S = \frac{π^{2}}{12}$ $S = - \frac{π^{2}}{12}$
	Commented by mathmax by abdo last updated on 25/Aug/20

	$sir shikari S is negative .!$
	Answered by mathmax by abdo last updated on 25/Aug/20
	$1)f(x) =u(x)+v(x) with u(x)=x^2 (even) and v =πx(odd) u(x) =(a_o /2) +Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T)∫_([T]) u(x)cos(nx)dx =(1/π)∫_(−π) ^π x^2 cos(nx) =(2/π)∫_0 ^π x^2 cos(nx)dx ⇒ (π/2)a_n =[(x^2 /n)sin(nx)]_0 ^π −∫_0 ^π ((2x)/n) sin(nx)dx =−(2/n)∫_0 ^π xsin(nx)dx =−(2/n){ [−(x/n)cos(nx)]_0 ^π +∫_0 ^π (1/n)cos(nx)dx} =−(2/n){ −(π/n)(−1)^n +(1/n^2 )[sin(nx)]_0 ^π } =((2π)/n^2 )(−1)^(n ) ⇒ a_n =((2π)/n^2 )×(2/π)(−1)^n =(4/n^2 )(−1)^n we hsve a_o =(2/π)∫_0 ^π x^2 dx =(2/π)[(x^3 /3)]_0 ^π =(2/π)×(π^3 /3) =((2π^2 )/3) ⇒(a_o /2) =(π^2 /3) ⇒x^2 =(π^2 /3) +4Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx) v(x) =πx =Σ_(n=1) ^∞ b_n sin(nx) ⇒ b_n =(2/T)∫_([T]) v(x)sin(nx)dx =(1/π)∫_(−π) ^π πxsin(nx)dx =2 ∫_0 ^π x sin(nx)dx =2{ [−(x/n)cos(nx)]_0 ^π +∫_0 ^π (1/n)cos)nx)dx} =2{−(π/n)(−1)^n +(1/n^2 )[sin(nx)]_0 ^π } =((−2π)/n) (−1)^n ⇒ v(x) =πx =−2π Σ_(n=1) ^∞ (((−1)^n )/n)sin(nx) ⇒ f(x) =(π^2 /3)+4Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx)−2π Σ_(n=1) ^∞ (((−1)^n )/n)sin(nx) 2) x=0 ⇒0 =(π^2 /3) +4 Σ_(n=1) ^∞ (((−1)^n )/n^2 ) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−(π^2 /(12))$
	$1) f (x) = u (x) + v (x) with u (x) = x^{2} (even) and v = π x (odd)$ $u (x) = \frac{a_{o}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (nx)$ $a_{n} = \frac{2}{T} \int_{[T]} u (x) \cos (nx) dx = \frac{1}{π} \int_{- π}^{π} x^{2} \cos (nx) = \frac{2}{π} \int_{0}^{π} x^{2} \cos (nx) dx \Rightarrow$ $\frac{π}{2} a_{n} = {[\frac{x^{2}}{n} \sin (nx)]}_{0}^{π} - \int_{0}^{π} \frac{2 x}{n} \sin (nx) dx = - \frac{2}{n} \int_{0}^{π} xsin (nx) dx$ $= - \frac{2}{n} {{[- \frac{x}{n} \cos (nx)]}_{0}^{π} + \int_{0}^{π} \frac{1}{n} \cos (nx) dx}$ $= - \frac{2}{n} {- \frac{π}{n} {(- 1)}^{n} + \frac{1}{n^{2}} {[\sin (nx)]}_{0}^{π}} = \frac{2 π}{n^{2}} {(- 1)}^{n} \Rightarrow$ $a_{n} = \frac{2 π}{n^{2}} \times \frac{2}{π} {(- 1)}^{n} = \frac{4}{n^{2}} {(- 1)}^{n} we hsve a_{o} = \frac{2}{π} \int_{0}^{π} x^{2} dx$ $= \frac{2}{π} {[\frac{x^{3}}{3}]}_{0}^{π} = \frac{2}{π} \times \frac{π^{3}}{3} = \frac{2 π^{2}}{3} \Rightarrow \frac{a_{o}}{2} = \frac{π^{2}}{3} \Rightarrow x^{2} = \frac{π^{2}}{3} + 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (nx)$ $v (x) = π x = \sum_{n = 1}^{\infty} b_{n} \sin (nx) \Rightarrow$ $b_{n} = \frac{2}{T} \int_{[T]} v (x) \sin (nx) dx = \frac{1}{π} \int_{- π}^{π} π xsin (nx) dx$ $= 2 \int_{0}^{π} x \sin (nx) dx = 2 {{[- \frac{x}{n} \cos (nx)]}_{0}^{π} + \int_{0}^{π} \frac{1}{n} \cos) nx) dx}$ $= 2 {- \frac{π}{n} {(- 1)}^{n} + \frac{1}{n^{2}} {[\sin (nx)]}_{0}^{π}} = \frac{- 2 π}{n} {(- 1)}^{n} \Rightarrow$ $v (x) = π x = - 2 π \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \sin (nx) \Rightarrow$ $f (x) = \frac{π^{2}}{3} + 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (nx) - 2 π \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \sin (nx)$ $2) x = 0 \Rightarrow 0 = \frac{π^{2}}{3} + 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = - \frac{π^{2}}{12}$
	Commented by Ar Brandon last updated on 25/Aug/20
	Je vous remercie��
	Commented by mathmax by abdo last updated on 25/Aug/20

	$you are welcome$
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