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Question Number 42266 by maxmathsup by imad last updated on 21/Aug/18
find the value of Σ_(n =0) ^∞ (1/(n^2 +1)) and Σ_(n=0) ^∞ (((−1)^n )/(n^2 +1)) .
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}\:=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:. \\ $$
Commented by maxmathsup by imad last updated on 22/Aug/18
we have proved that e^(−∣x∣) =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ ((1−(−1)^n e^(−π) )/(n^2 +1)) cos(nx) x=0 ⇒1 = ((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ (1/(n^2 +1)) −((2e^(−π) )/π) Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) let s =Σ_(n=1) ^∞ (1/(n^2 +1)) and w =Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) ⇒ 1−((1−e^(−π) )/π) =(2/π) s −(2/π)e^(−π) w ⇒(π/2)(((π−1+e^(−π) )/π))=s−e^(−π) w ⇒ s −e^(−π) w =((π−1+e^(−π) )/2) x =π ⇒e^(−π) =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) −((2e^(−π) )/π) Σ_(n=1) ^∞ (1/(n^2 +1)) ⇒ e^(−π) −((1−e^(−π) )/π) =(2/π)w −((2e^(−π) )/π) s ⇒(π/2)(((π e^(−π) −1 +e^(−π) )/π)) =−2e^(−π) s +w ⇒ ((π e^(−π) −1+e^(−π) )/2) =−2 e^(−π) s +w we get the system s −e^(−π) w =((π−1 +e^(−π) )/2) and −2 e^(−π) s +w =(((π+1)e^(−π) )/2) ⇒ −2 e^(−2π) s +e^(−π) w =(((π+1)e^(−2π) )/2) ⇒ (1−2e^(−2π) )s = ((π−1+e^(−π) +(π+1)e^(−2π) )/2) ⇒ s =((π−1 +e^(−π) +(π+1)e^(−2π) )/(2(1−2 e^(−2π) ))) and w =(((π+1)e^(−π) )/2) +2 e^(−π) s .
$${we}\:{have}\:{proved}\:{that}\: \\ $$$${e}^{−\mid{x}\mid} \:=\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:\:+\frac{\mathrm{2}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:{cos}\left({nx}\right) \\ $$$${x}=\mathrm{0}\:\Rightarrow\mathrm{1}\:=\:\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{2}{e}^{−\pi} }{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} +\mathrm{1}}\:\:{let}\: \\ $$$${s}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:{and}\:{w}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{1}−\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:=\frac{\mathrm{2}}{\pi}\:{s}\:−\frac{\mathrm{2}}{\pi}{e}^{−\pi} \:{w}\:\Rightarrow\frac{\pi}{\mathrm{2}}\left(\frac{\pi−\mathrm{1}+{e}^{−\pi} }{\pi}\right)={s}−{e}^{−\pi} \:{w}\:\Rightarrow \\ $$$${s}\:−{e}^{−\pi} {w}\:=\frac{\pi−\mathrm{1}+{e}^{−\pi} }{\mathrm{2}} \\ $$$${x}\:=\pi\:\Rightarrow{e}^{−\pi} \:=\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{2}{e}^{−\pi} }{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${e}^{−\pi} \:−\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:=\frac{\mathrm{2}}{\pi}{w}\:−\frac{\mathrm{2}{e}^{−\pi} }{\pi}\:{s}\:\Rightarrow\frac{\pi}{\mathrm{2}}\left(\frac{\pi\:{e}^{−\pi} \:−\mathrm{1}\:+{e}^{−\pi} }{\pi}\right)\:=−\mathrm{2}{e}^{−\pi} \:{s}\:+{w}\:\Rightarrow \\ $$$$\frac{\pi\:{e}^{−\pi} \:−\mathrm{1}+{e}^{−\pi} }{\mathrm{2}}\:\:=−\mathrm{2}\:{e}^{−\pi} {s}\:+{w}\:\:{we}\:{get}\:{the}\:{system} \\ $$$$\boldsymbol{\mathrm{s}}\:−\boldsymbol{\mathrm{e}}^{−\pi} {w}\:=\frac{\pi−\mathrm{1}\:+{e}^{−\pi} }{\mathrm{2}}\:\:{and}\:\:−\mathrm{2}\:{e}^{−\pi} \:{s}\:+{w}\:=\frac{\left(\pi+\mathrm{1}\right){e}^{−\pi} }{\mathrm{2}}\:\Rightarrow \\ $$$$−\mathrm{2}\:{e}^{−\mathrm{2}\pi} \:{s}\:+{e}^{−\pi} {w}\:=\frac{\left(\pi+\mathrm{1}\right){e}^{−\mathrm{2}\pi} }{\mathrm{2}}\:\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{2}{e}^{−\mathrm{2}\pi} \right){s}\:=\:\frac{\pi−\mathrm{1}+{e}^{−\pi} \:\:+\left(\pi+\mathrm{1}\right){e}^{−\mathrm{2}\pi} }{\mathrm{2}}\:\Rightarrow \\ $$$${s}\:=\frac{\pi−\mathrm{1}\:+{e}^{−\pi} \:+\left(\pi+\mathrm{1}\right){e}^{−\mathrm{2}\pi} }{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\:{e}^{−\mathrm{2}\pi} \right)}\:\:\:\:{and}\:\:{w}\:=\frac{\left(\pi+\mathrm{1}\right){e}^{−\pi} }{\mathrm{2}}\:+\mathrm{2}\:{e}^{−\pi} \:{s}\:. \\ $$

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