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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=Σ_(n=1) ^∞ (((−1)^n )/n^2 )=(−1)Σ_(n=1) ^∞ (((−1)^(n+1) )/n^2 )=(−1)((1/1)−(1/2^2 )+(1/3^2 )−(1/4^2 )+.....)  S_n =Σ_(n=1) ^∞ (1/n^2 )=1+(1/2^2 )+(1/3^2 )+....  S_n −S=2((1/2^2 )+(1/4^2 )+....)=2Σ_(n=1) ^∞ (1/((2n)^2 ))  Σ^∞ (1/n^2 )−Σ^∞ (1/(2n^2 ))=S  (1/2)Σ^∞ (1/n^2 )=S  (1/2).(π^2 /6)=S  S=(π^2 /(12))    S=−(π^2 /(12))

$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\left(−\mathrm{1}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{2}} }=\left(−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+.....\right) \\ $$$${S}_{{n}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+.... \\ $$$${S}_{{n}} −{S}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+....\right)=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} } \\ $$$$\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\overset{\infty} {\sum}\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }={S} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }={S} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}={S} \\ $$$${S}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$ \\ $$$${S}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

Commented by mathmax by abdo last updated on 25/Aug/20

sir shikari S is negative.!

$$\mathrm{sir}\:\mathrm{shikari}\:\mathrm{S}\:\mathrm{is}\:\mathrm{negative}.! \\ $$

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))

$$\left.\mathrm{1}\right)\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{u}\left(\mathrm{x}\right)+\mathrm{v}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{u}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \left(\mathrm{even}\right)\:\mathrm{and}\:\mathrm{v}\:=\pi\mathrm{x}\left(\mathrm{odd}\right) \\ $$$$\mathrm{u}\left(\mathrm{x}\right)\:=\frac{\mathrm{a}_{\mathrm{o}} }{\mathrm{2}}\:+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}_{\mathrm{n}} \:\mathrm{cos}\left(\mathrm{nx}\right) \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{2}}{\mathrm{T}}\int_{\left[\mathrm{T}\right]} \mathrm{u}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\pi} \mathrm{x}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{nx}\right)\:=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \:\mathrm{x}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{a}_{\mathrm{n}} =\left[\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}}\mathrm{sin}\left(\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\pi} −\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2x}}{\mathrm{n}}\:\mathrm{sin}\left(\mathrm{nx}\right)\mathrm{dx}\:=−\frac{\mathrm{2}}{\mathrm{n}}\int_{\mathrm{0}} ^{\pi} \:\mathrm{xsin}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{n}}\left\{\:\:\left[−\frac{\mathrm{x}}{\mathrm{n}}\mathrm{cos}\left(\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\pi} +\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{n}}\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}\right\} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{n}}\left\{\:−\frac{\pi}{\mathrm{n}}\left(−\mathrm{1}\right)^{\mathrm{n}} \:+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\left[\mathrm{sin}\left(\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\pi} \right\}\:=\frac{\mathrm{2}\pi}{\mathrm{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{\mathrm{n}\:} \:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{2}\pi}{\mathrm{n}^{\mathrm{2}} }×\frac{\mathrm{2}}{\pi}\left(−\mathrm{1}\right)^{\mathrm{n}} \:=\frac{\mathrm{4}}{\mathrm{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{\mathrm{n}} \:\:\mathrm{we}\:\mathrm{hsve}\:\mathrm{a}_{\mathrm{o}} =\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \:\mathrm{x}^{\mathrm{2}} \:\mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\left[\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\pi} \:=\frac{\mathrm{2}}{\pi}×\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}}\:\Rightarrow\frac{\mathrm{a}_{\mathrm{o}} }{\mathrm{2}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:+\mathrm{4}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{nx}\right) \\ $$$$\mathrm{v}\left(\mathrm{x}\right)\:=\pi\mathrm{x}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{b}_{\mathrm{n}} \mathrm{sin}\left(\mathrm{nx}\right)\:\Rightarrow \\ $$$$\mathrm{b}_{\mathrm{n}} =\frac{\mathrm{2}}{\mathrm{T}}\int_{\left[\mathrm{T}\right]} \:\mathrm{v}\left(\mathrm{x}\right)\mathrm{sin}\left(\mathrm{nx}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\pi} \pi\mathrm{xsin}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$$$\left.=\left.\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\mathrm{x}\:\mathrm{sin}\left(\mathrm{nx}\right)\mathrm{dx}\:=\mathrm{2}\left\{\:\left[−\frac{\mathrm{x}}{\mathrm{n}}\mathrm{cos}\left(\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\pi} +\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{n}}\mathrm{cos}\right)\mathrm{nx}\right)\mathrm{dx}\right\} \\ $$$$=\mathrm{2}\left\{−\frac{\pi}{\mathrm{n}}\left(−\mathrm{1}\right)^{\mathrm{n}} \:+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\left[\mathrm{sin}\left(\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\pi} \right\}\:=\frac{−\mathrm{2}\pi}{\mathrm{n}}\:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\Rightarrow \\ $$$$\mathrm{v}\left(\mathrm{x}\right)\:=\pi\mathrm{x}\:=−\mathrm{2}\pi\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\mathrm{sin}\left(\mathrm{nx}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}+\mathrm{4}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{nx}\right)−\mathrm{2}\pi\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\mathrm{sin}\left(\mathrm{nx}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{x}=\mathrm{0}\:\Rightarrow\mathrm{0}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:+\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$ \\ $$

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

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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