Question Number 82349 by Tony Lin last updated on 20/Feb/20
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=? \\ $$
Answered by mind is power last updated on 21/Feb/20
![we can observe that (1/(1+n^2 ))=Re{(1/(1−in))} ∫_0 ^π e^x cos(−nx)dx=Re{∫_0 ^π e^x e^(−inx) dx} =Re{∫_0 ^π e^(x(1−in)) dx} =Re{((e^π (−1)^n −1)/(1−in))}=((e^π (−1)^n −1)/(1+n^2 )) not what we want but near let f(x)=e^(∣x∣) x∈[−π ,π]∩C_0 2π periodique since f is odd fourier serie of f is S_n (f)=a_0 +Σ_(k≥1) a_k (f)cos(kx) a_0 =(1/(2π))∫_(−π) ^π e^(∣x∣) dx=(1/π)∫_0 ^π e^x dx=((e^π −1)/π) a_k =(1/π)∫_(−π) ^π e^(∣x∣) cos(kx)dx=(2/π)∫_0 ^π e^x cos(kx)dx =(2/π)((e^π (−1)^k −1)/(k^2 +1)) S_n (f)=((e^π −1)/π)+(2/π)Σ_(k≥1) (((−1)^k e^π −1)/(k^2 +1))cos(kx)=f(x)^ ∀∈[−π,π] for x=0 1=((e^π −1)/π)+(2/π)Σ_(k≥1) (((−1)^k e^π −1)/(k^2 +1)) x=π⇒e^π =((e^π −1)/π)+(2/π)Σ_(k≥1) (((−1)^k e^π −1)/(k^2 +1))(−1)^k S=Σ_(k≥1) (1/(1+k^2 )),T=Σ_(k≥1) (((−1)^k )/(1+k^2 )) ⇒1=((e^π −1)/π)+((2e^π )/π)T−(2/π)S e^π =((e^π −1)/π)+(2/π)e^π S−(2/π)T ⇒e^(2π) +1=((e^(2π) −1)/π)+(2/π)(e^(2π) −1)S ⇒S=(π/2).((e^(2π) +1)/(e^(2π) −1))−(1/2) S=(1/2)(πcoth(π)−1)=Σ_(k≥1) (1/(1+k^2 ))](https://www.tinkutara.com/question/Q82493.png)
$${we}\:{can} \\ $$$${observe}\:\:{that}\:\frac{\mathrm{1}}{\mathrm{1}+{n}^{\mathrm{2}} }={Re}\left\{\frac{\mathrm{1}}{\mathrm{1}−{in}}\right\} \\ $$$$\int_{\mathrm{0}} ^{\pi} {e}^{{x}} {cos}\left(−{nx}\right){dx}={Re}\left\{\int_{\mathrm{0}} ^{\pi} {e}^{{x}} {e}^{−{inx}} {dx}\right\} \\ $$$$={Re}\left\{\int_{\mathrm{0}} ^{\pi} {e}^{{x}\left(\mathrm{1}−{in}\right)} {dx}\right\} \\ $$$$={Re}\left\{\frac{{e}^{\pi} \left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{\mathrm{1}−{in}}\right\}=\frac{{e}^{\pi} \left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{\mathrm{1}+{n}^{\mathrm{2}} }\:\:{not}\:{what}\:{we}\:{want}\:{but}\:{near} \\ $$$${let}\:{f}\left({x}\right)={e}^{\mid{x}\mid} \:{x}\in\left[−\pi\:\:,\pi\right]\cap{C}_{\mathrm{0}} \:\mathrm{2}\pi\:{periodique}\:{since}\:{f}\:{is}\:{odd} \\ $$$${fourier}\:{serie}\:{of}\:{f}\:{is}\: \\ $$$${S}_{{n}} \left({f}\right)={a}_{\mathrm{0}} +\underset{{k}\geqslant\mathrm{1}} {\sum}{a}_{{k}} \left({f}\right){cos}\left({kx}\right) \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} {e}^{\mid{x}\mid} {dx}=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {e}^{{x}} {dx}=\frac{{e}^{\pi} −\mathrm{1}}{\pi} \\ $$$${a}_{{k}} =\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\pi} {e}^{\mid{x}\mid} {cos}\left({kx}\right){dx}=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} {e}^{{x}} {cos}\left({kx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\frac{{e}^{\pi} \left(−\mathrm{1}\right)^{{k}} −\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{1}} \\ $$$${S}_{{n}} \left({f}\right)=\frac{{e}^{\pi} −\mathrm{1}}{\pi}+\frac{\mathrm{2}}{\pi}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {e}^{\pi} −\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{1}}{cos}\left({kx}\right)={f}\left({x}\bar {\right)}\forall\in\left[−\pi,\pi\right] \\ $$$${for}\:{x}=\mathrm{0} \\ $$$$\mathrm{1}=\frac{{e}^{\pi} −\mathrm{1}}{\pi}+\frac{\mathrm{2}}{\pi}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {e}^{\pi} −\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{1}} \\ $$$${x}=\pi\Rightarrow{e}^{\pi} =\frac{{e}^{\pi} −\mathrm{1}}{\pi}+\frac{\mathrm{2}}{\pi}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {e}^{\pi} −\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{1}}\left(−\mathrm{1}\right)^{{k}} \\ $$$${S}=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{k}^{\mathrm{2}} },{T}=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}=\frac{{e}^{\pi} −\mathrm{1}}{\pi}+\frac{\mathrm{2}{e}^{\pi} }{\pi}{T}−\frac{\mathrm{2}}{\pi}{S} \\ $$$${e}^{\pi} =\frac{{e}^{\pi} −\mathrm{1}}{\pi}+\frac{\mathrm{2}}{\pi}{e}^{\pi} {S}−\frac{\mathrm{2}}{\pi}{T} \\ $$$$\Rightarrow{e}^{\mathrm{2}\pi} +\mathrm{1}=\frac{{e}^{\mathrm{2}\pi} −\mathrm{1}}{\pi}+\frac{\mathrm{2}}{\pi}\left({e}^{\mathrm{2}\pi} −\mathrm{1}\right){S} \\ $$$$\Rightarrow{S}=\frac{\pi}{\mathrm{2}}.\frac{{e}^{\mathrm{2}\pi} +\mathrm{1}}{{e}^{\mathrm{2}\pi} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi{coth}\left(\pi\right)−\mathrm{1}\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$