calculateA_n = Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(n) andB_n = Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(((nπ)/3))


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Question Number 132496 by mathmax by abdo last updated on 14/Feb/21

${calculateA}_{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (n) {andB}_{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (\frac{n π}{3})$
	Answered by mnjuly1970 last updated on 14/Feb/21
	$by using dilogarithm function or fourier series... first method.. f(x)=x x∈[−π,π] f is artificial periodic function on [−π ,π] f(x)≈(a_0 /2)+Σ_(n=1) (a_n cos(nx)+b_n sin(nx)) a_n =(1/π)∫_(−π) ^( π) f(x)cos(nx)dx=0 b_n =(1/π)∫_(−π) ^( π) f(x)sin(nx)dx =(1/π)∫_(−π) ^( π) xsin(nx)dx=(2/π)∫_0 ^( π) xsin(nx)dx =(2/π){[−(x/n)cos(nx)]_(0 ) ^π +(1/n)∫_0 ^( π) cos(nx)dx 2(−1)^(n+1) (1/n) a_0 =(1/π)∫_(−π) ^( π) f(x)dx=0 x≈ Σ_(n=1) ^∞ ((2(−1)^(n+1) )/(n ))sin(nx) ∫xdx≈2Σ_(n=1) ^∞ ∫ (((−1)^(n+1) )/(n ))sin(nx)dx (x^2 /2)≈C+2Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx) x=π⇒ (π^2 /2)≈C+2Σ_(n=1) (1/n^2 ) C=(π^2 /2)−(π^2 /3)=(π^2 /6) (x^2 /2)=(π^2 /6)+2Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx) x=1 ⇒(1/2)−(π^2 /6)=2Σ_(n=1) ^∞ (((−1)^n )/n^2 ) Σ_(n=1) ^∞ (((−1)^n )/n^2 )=(1/4)−(π^2 /(12)) ...$
	$b y u s i n g d i l o g a r i t h m f u n c t i o n$ $o r f o u r i e r s e r i e s . . .$ $f i r s t m e t h o d . .$ $f (x) = x x \in [- π, π]$ $f i s a r t i f i c i a l p e r i o d i c f u n c t i o n$ $o n [- π, π]$ $f (x) \approx \frac{a_{0}}{2} + \sum_{n = 1} (a_{n} c o s (n x) + b_{n} s i n (n x))$ $a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) c o s (n x) d x = 0$ $b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) s i n (n x) d x$ $= \frac{1}{π} \int_{- π}^{π} x s i n (n x) d x = \frac{2}{π} \int_{0}^{π} x s i n (n x) d x$ $= \frac{2}{π} {{[- \frac{x}{n} c o s (n x)]}_{0}^{π} + \frac{1}{n} \int_{0}^{π} c o s (n x) d x$ $2 {(- 1)}^{n + 1} \frac{1}{n}$ $a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = 0$ $x \approx \underset{n = 1}{\sum^{\infty}} \frac{2 {(- 1)}^{n + 1}}{n} s i n (n x)$ $\int x d x \approx 2 \underset{n = 1}{\sum^{\infty}} \int \frac{{(- 1)}^{n + 1}}{n} s i n (n x) d x$ $\frac{x^{2}}{2} \approx C + 2 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x)$ $x = π \Rightarrow \frac{π^{2}}{2} \approx C + 2 \sum_{n = 1} \frac{1}{n^{2}}$ $C = \frac{π^{2}}{2} - \frac{π^{2}}{3} = \frac{π^{2}}{6}$ $\frac{x^{2}}{2} = \frac{π^{2}}{6} + 2 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x)$ $x = 1 \Rightarrow \frac{1}{2} - \frac{π^{2}}{6} = 2 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}}$ $\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} = \frac{1}{4} - \frac{π^{2}}{12} . . .$
	Commented by mathmax by abdo last updated on 14/Feb/21

	$thank you sir .$
	Commented by mnjuly1970 last updated on 15/Feb/21

	$g r a t e f u l . . .$
	Answered by mnjuly1970 last updated on 14/Feb/21

	$s e c o n d m e t h o d$ $\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} (\frac{e^{i n} + e^{- i n}}{2})$ $= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- e^{i})}^{n}}{n^{2}} + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- e^{- i})}^{n}}{n^{2}}$ $= \frac{1}{2} ({l i}_{2} (- e^{i}) + {l i}_{2} (- \frac{1}{e^{i}}))$ $\underset{\frac{- π^{2}}{6} - \frac{1}{2} {l n}^{2} (z)}{\overset{{l i}_{2} (- z) + {l i}_{2} (- \frac{1}{z})}{=}} \frac{1}{2} (\frac{- π^{2}}{6} - \frac{1}{2} {l n}^{2} (e^{i}))$ $= \frac{- π^{2}}{12} - \frac{1}{4} (i^{2}) = \frac{- π^{2}}{12} + \frac{1}{4} . . . .$
	Answered by mathmax by abdo last updated on 14/Feb/21

	$let φ (x) = x, 2 π periodic odd \Rightarrow φ (x) = \sum_{n = 1}^{\infty} a_{n} \sin (nx)$ $a_{n} = \frac{1}{π} \int_{- π}^{π} x \sin (nx) dx = \frac{2}{π} \int_{0}^{π} xsin (nx) dx \Rightarrow$ $\frac{π}{2} a_{n} = {[- \frac{x}{n} \cos (nx)]}_{0}^{π} + \int_{0}^{π} \frac{1}{n} \cos (nx) dx = - \frac{π}{n} {(- 1)}^{n} + \frac{1}{n^{2}} {[\sin (nx)]}_{0}^{π}$ $= - \frac{π}{n} {(- 1)}^{n} \Rightarrow a_{n} = \frac{2}{π} . (- \frac{π}{n}) {(- 1)}^{n} = - \frac{2}{n} {(- 1)}^{n} \Rightarrow$ $x = - 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \sin (nx) \Rightarrow \frac{x^{2}}{2} = 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (nx) + C$ $x = 0 \Rightarrow 0 = 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} + c = 2 . (2^{1 - 2} - 1) ξ (2) + C$ $= 2 (- \frac{1}{2}) \frac{π^{2}}{6} + C = - \frac{π^{2}}{6} + C \Rightarrow C = \frac{π^{2}}{6} \Rightarrow$ $\frac{x^{2}}{2} = 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (nx) + \frac{π^{2}}{6} \Rightarrow 2 Σ (. . . .) = \frac{x^{2}}{2} - \frac{π^{2}}{6} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (nx) = \frac{x^{2}}{4} - \frac{π^{2}}{12}$ $x = 1 \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (n) = \frac{1}{4} - \frac{π^{2}}{12}$ $x = \frac{π}{3} \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (n \frac{π}{3}) = \frac{π^{2}}{36} - \frac{π^{2}}{12} = \frac{π^{2}}{36} - \frac{3 π^{2}}{36} = - \frac{π^{2}}{18}$
	Commented by mnjuly1970 last updated on 15/Feb/21

	$v e r y n i c e s i r m a x . .$
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