Question Number 121326 by Dwaipayan Shikari last updated on 06/Nov/20
$$\frac{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}}{\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}} \\ $$
Commented by Dwaipayan Shikari last updated on 06/Nov/20
![Σ_(n=2) ^∞ (1/(n^2 −1))=(1/2)(Σ_(n=2) ^∞ (1/(n−1))−Σ_(n=2) ^∞ (1/(n+1)))=(1/2)(1+(1/2)+(1/3)+...−(1/3)−(1/4)−...) =(3/4) Σ_(n=1) ^∞ (1/(n^2 +1))=(1/(2i))Σ_(n=1) ^∞ (1/(n−i))−(1/(n+i)) =(1/(2i))Σ_(n=2) ^∞ ((1/n)−(1/(n−i−1)))−(1/(2i))Σ_(n=2) ^∞ ((1/n)−(1/(n+i−1))) =(1/(2i))(ψ(i)−ψ(1−i)) =(1/(2i))(−πcot(iπ))+i) = (1/(2i))(−πcot(iπ)+1))=(π/2)cothπ+(1/2) [Γ(s)Γ(1−s)=(π/(sinπs))] [log(Γ(s)+log(Γ(1−s))=logπ−log(sinπs)] [((Γ′(s))/(Γ(s)))−((Γ′(1−s))/(Γ(1−s)))=−((πcossπs)/(sinπs))] [𝛙(1−s)−𝛙(s)=πcotπs]](https://www.tinkutara.com/question/Q121331.png)
$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\mathrm{1}}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}−…\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}{i}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−{i}}−\frac{\mathrm{1}}{{n}+{i}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−{i}−\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}{i}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+{i}−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\psi\left({i}\right)−\psi\left(\mathrm{1}−{i}\right)\right) \\ $$$$\left.=\left.\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−\pi{cot}\left({i}\pi\right)\right)+{i}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−\pi{cot}\left({i}\pi\right)+\mathrm{1}\right)\right)=\frac{\pi}{\mathrm{2}}{coth}\pi+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left[\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\frac{\pi}{{sin}\pi{s}}\right] \\ $$$$\left[{log}\left(\Gamma\left({s}\right)+{log}\left(\Gamma\left(\mathrm{1}−{s}\right)\right)={log}\pi−{log}\left({sin}\pi{s}\right)\right]\right. \\ $$$$\left[\frac{\Gamma'\left({s}\right)}{\Gamma\left({s}\right)}−\frac{\Gamma'\left(\mathrm{1}−{s}\right)}{\Gamma\left(\mathrm{1}−{s}\right)}=−\frac{\pi{coss}\pi{s}}{{sin}\pi{s}}\right] \\ $$$$\left[\boldsymbol{\psi}\left(\mathrm{1}−{s}\right)−\boldsymbol{\psi}\left({s}\right)=\pi{cot}\pi{s}\right] \\ $$$$ \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 06/Nov/20
$${Is}\:{it}\:{right}? \\ $$
Commented by mathmax by abdo last updated on 06/Nov/20
$$\mathrm{perhaps}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}..! \\ $$
Answered by mathmax by abdo last updated on 06/Nov/20
![let f(x) =e^(−∣x∣) function 2π periodic even we have f(x) =(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n =(2/T)∫_T f(x)cos(nx)dx =(2/(2π))∫_(−π) ^π e^(−∣x∣) cos(nx)dx =(2/π)∫_0 ^π e^(−x) cos(nx)dx ⇒(π/2) a_n = Re(∫_0 ^π e^(−x+inx) dx) and ∫_0 ^π e^((−1+in)x) dx =[(1/(−1+in)) e^((−1+in)x) ]_0 ^π =((−1)/(1−in)){ e^((−1+in)π) −1} =−((1+in)/(1+n^2 )){(−1)^n e^(−π) −1} =((1−(−1)^n e^(−π) )/(1+n^2 )) +i(...) ⇒ a_n =(2/π)×((1−(−1)^n e^(−π) )/(1+n^2 )) ⇒ a_0 =(2/π)(1−e^(−π) ) ⇒(a_0 /2)=((1−e^(−π) )/π) ⇒ e^(−∣x∣) =((1−e^(−π) )/π) +(2/π)Σ_(n=1) ^∞ ((1−(−1)^n e^(−π) )/(1+n^2 )) cos(nx) ⇒ x=0 ⇒1 =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ (1/(1+n^2 )) −((2e^(−π) )/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) ⇒1 =((1−e^(−π) )/π) +(2/π)x −((2e^(−π) )/π)y x=π ⇒e^(−π) =((1−e^(−π) )/2) +(2/π)Σ_(n=1) ^∞ (−1)^n ((1−(−1)^n e^(−π) )/(n^2 +1)) ⇒ e^(−π) =((1−e^(−π) )/2) +(2/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1))−((2e^(−π) )/π)Σ_(n=1) ^∞ (1/(n^2 +1)) =((1−e^(−π) )/2) +(2/π)y−((2e^(−π) )/π) x we get the system { (((2/π)x−((2e^(−π) )/π)y =1−((1−e^(−π) )/π) ⇒)),((−((2e^(−π) )/π)x+(2/π)y =e^(−π) −((1−e^(−π) )/2))) :} { ((2x−2e^(−π) y =π−1+e^(−π) )),((−2e^(−π) x +2y =πe^(−π) −((π−πe^(−π) )/2))) :} Δ_s =4−4e^(−2π) ⇒x=(Δ_x /Δ) Δ_x = determinant (((π−1+e^(−π) −2e^(−π) )),((πe^(−π) −((π−πe^(−π) )/2) 2)))=.... x =Σ_(n=1) ^∞ (1/(n^2 +1)) ....be continued...](https://www.tinkutara.com/question/Q121347.png)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{−\mid\mathrm{x}\mid} \:\mathrm{function}\:\mathrm{2}\pi\:\mathrm{periodic}\:\mathrm{even}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}_{\mathrm{n}} \mathrm{cos}\left(\mathrm{nx}\right)\:\:\mathrm{with}\:\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{2}}{\mathrm{T}}\int_{\mathrm{T}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} \:\mathrm{e}^{−\mid\mathrm{x}\mid} \:\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}\:=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \:\mathrm{e}^{−\mathrm{x}} \:\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$$$\Rightarrow\frac{\pi}{\mathrm{2}}\:\mathrm{a}_{\mathrm{n}} =\:\mathrm{Re}\left(\int_{\mathrm{0}} ^{\pi} \:\mathrm{e}^{−\mathrm{x}+\mathrm{inx}} \mathrm{dx}\right)\:\mathrm{and}\:\int_{\mathrm{0}} ^{\pi} \:\mathrm{e}^{\left(−\mathrm{1}+\mathrm{in}\right)\mathrm{x}} \mathrm{dx} \\ $$$$=\left[\frac{\mathrm{1}}{−\mathrm{1}+\mathrm{in}}\:\mathrm{e}^{\left(−\mathrm{1}+\mathrm{in}\right)\mathrm{x}} \right]_{\mathrm{0}} ^{\pi} \:=\frac{−\mathrm{1}}{\mathrm{1}−\mathrm{in}}\left\{\:\mathrm{e}^{\left(−\mathrm{1}+\mathrm{in}\right)\pi} −\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{1}+\mathrm{in}}{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\left\{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\pi} −\mathrm{1}\right\}\:=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{−\pi} }{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:+\mathrm{i}\left(…\right)\:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{2}}{\pi}×\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\pi} }{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{0}} =\frac{\mathrm{2}}{\pi}\left(\mathrm{1}−\mathrm{e}^{−\pi} \right)\:\Rightarrow\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{2}}=\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\pi}\:\Rightarrow \\ $$$$\mathrm{e}^{−\mid\mathrm{x}\mid} \:=\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{−\pi} }{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:\mathrm{cos}\left(\mathrm{nx}\right)\:\Rightarrow \\ $$$$\mathrm{x}=\mathrm{0}\:\Rightarrow\mathrm{1}\:=\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:−\frac{\mathrm{2e}^{−\pi} }{\pi}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{1}\:=\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\mathrm{x}\:−\frac{\mathrm{2e}^{−\pi} }{\pi}\mathrm{y} \\ $$$$\mathrm{x}=\pi\:\Rightarrow\mathrm{e}^{−\pi} \:=\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\mathrm{2}}\:+\frac{\mathrm{2}}{\pi}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{−\pi} }{\mathrm{n}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{e}^{−\pi} \:=\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\mathrm{2}}\:+\frac{\mathrm{2}}{\pi}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{2e}^{−\pi} }{\pi}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\mathrm{2}}\:+\frac{\mathrm{2}}{\pi}\mathrm{y}−\frac{\mathrm{2e}^{−\pi} }{\pi}\:\mathrm{x}\:\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{\frac{\mathrm{2}}{\pi}\mathrm{x}−\frac{\mathrm{2e}^{−\pi} }{\pi}\mathrm{y}\:=\mathrm{1}−\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\pi}\:\:\:\:\:\Rightarrow}\\{−\frac{\mathrm{2e}^{−\pi} }{\pi}\mathrm{x}+\frac{\mathrm{2}}{\pi}\mathrm{y}\:=\mathrm{e}^{−\pi} −\frac{\mathrm{1}−\mathrm{e}^{−\pi} }{\mathrm{2}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2x}−\mathrm{2e}^{−\pi} \mathrm{y}\:=\pi−\mathrm{1}+\mathrm{e}^{−\pi} }\\{−\mathrm{2e}^{−\pi} \:\mathrm{x}\:+\mathrm{2y}\:=\pi\mathrm{e}^{−\pi} −\frac{\pi−\pi\mathrm{e}^{−\pi} }{\mathrm{2}}}\end{cases} \\ $$$$\Delta_{\mathrm{s}} =\mathrm{4}−\mathrm{4e}^{−\mathrm{2}\pi} \:\Rightarrow\mathrm{x}=\frac{\Delta_{\mathrm{x}} }{\Delta} \\ $$$$\Delta_{\mathrm{x}} =\begin{vmatrix}{\pi−\mathrm{1}+\mathrm{e}^{−\pi} \:\:\:\:\:\:\:\:\:\:−\mathrm{2e}^{−\pi} \:\:\:}\\{\pi\mathrm{e}^{−\pi} −\frac{\pi−\pi\mathrm{e}^{−\pi} }{\mathrm{2}}\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}=…. \\ $$$$\mathrm{x}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} \:+\mathrm{1}}\:\:\:….\mathrm{be}\:\mathrm{continued}… \\ $$