((Σ_(n=1) ^∞ (1/(n^2 +1)))/(Σ_(n=2) ^∞ (1/(n^2 −1))))


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Question Number 121326 by Dwaipayan Shikari last updated on 06/Nov/20

$\frac{\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 1}}{\underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2} - 1}}$
Commented by Dwaipayan Shikari last updated on 06/Nov/20

$\underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2} - 1} = \frac{1}{2} (\underset{n = 2}{\sum^{\infty}} \frac{1}{n - 1} - \underset{n = 2}{\sum^{\infty}} \frac{1}{n + 1}) = \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + . . . - \frac{1}{3} - \frac{1}{4} - . . .)$ $= \frac{3}{4}$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 1} = \frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} \frac{1}{n - i} - \frac{1}{n + i}$ $= \frac{1}{2 i} \underset{n = 2}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n - i - 1}) - \frac{1}{2 i} \underset{n = 2}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + i - 1})$ $= \frac{1}{2 i} (ψ (i) - ψ (1 - i))$ $= \frac{1}{2 i} (- π c o t (i π)) + i) = \frac{1}{2 i} (- π c o t (i π) + 1)) = \frac{π}{2} c o t h π + \frac{1}{2}$ $[Γ (s) Γ (1 - s) = \frac{π}{s i n π s}]$ $[l o g (Γ (s) + l o g (Γ (1 - s)) = l o g π - l o g (s i n π s)]$ $[\frac{Γ^{'} (s)}{Γ (s)} - \frac{Γ^{'} (1 - s)}{Γ (1 - s)} = - \frac{π c o s s π s}{s i n π s}]$ $[ψ (1 - s) - ψ (s) = π c o t π s]$
Commented by Dwaipayan Shikari last updated on 06/Nov/20

$I s i t r i g h t ?$
Commented by mathmax by abdo last updated on 06/Nov/20

$perhaps your answer is correct . .!$
	Answered by mathmax by abdo last updated on 06/Nov/20
	$let f(x) =e^(−∣x∣) function 2π periodic even we have f(x) =(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n =(2/T)∫_T f(x)cos(nx)dx =(2/(2π))∫_(−π) ^π e^(−∣x∣) cos(nx)dx =(2/π)∫_0 ^π e^(−x) cos(nx)dx ⇒(π/2) a_n = Re(∫_0 ^π e^(−x+inx) dx) and ∫_0 ^π e^((−1+in)x) dx =[(1/(−1+in)) e^((−1+in)x) ]_0 ^π =((−1)/(1−in)){ e^((−1+in)π) −1} =−((1+in)/(1+n^2 )){(−1)^n e^(−π) −1} =((1−(−1)^n e^(−π) )/(1+n^2 )) +i(...) ⇒ a_n =(2/π)×((1−(−1)^n e^(−π) )/(1+n^2 )) ⇒ a_0 =(2/π)(1−e^(−π) ) ⇒(a_0 /2)=((1−e^(−π) )/π) ⇒ e^(−∣x∣) =((1−e^(−π) )/π) +(2/π)Σ_(n=1) ^∞ ((1−(−1)^n e^(−π) )/(1+n^2 )) cos(nx) ⇒ x=0 ⇒1 =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ (1/(1+n^2 )) −((2e^(−π) )/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) ⇒1 =((1−e^(−π) )/π) +(2/π)x −((2e^(−π) )/π)y x=π ⇒e^(−π) =((1−e^(−π) )/2) +(2/π)Σ_(n=1) ^∞ (−1)^n ((1−(−1)^n e^(−π) )/(n^2 +1)) ⇒ e^(−π) =((1−e^(−π) )/2) +(2/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1))−((2e^(−π) )/π)Σ_(n=1) ^∞ (1/(n^2 +1)) =((1−e^(−π) )/2) +(2/π)y−((2e^(−π) )/π) x we get the system { (((2/π)x−((2e^(−π) )/π)y =1−((1−e^(−π) )/π) ⇒)),((−((2e^(−π) )/π)x+(2/π)y =e^(−π) −((1−e^(−π) )/2))) :} { ((2x−2e^(−π) y =π−1+e^(−π) )),((−2e^(−π) x +2y =πe^(−π) −((π−πe^(−π) )/2))) :} Δ_s =4−4e^(−2π) ⇒x=(Δ_x /Δ) Δ_x = determinant (((π−1+e^(−π) −2e^(−π) )),((πe^(−π) −((π−πe^(−π) )/2) 2)))=.... x =Σ_(n=1) ^∞ (1/(n^2 +1)) ....be continued...$
	$let f (x) = e^{- ∣ x ∣} function 2 π periodic even we have$ $f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (nx) with a_{n} = \frac{2}{T} \int_{T} f (x) \cos (nx) dx$ $= \frac{2}{2 π} \int_{- π}^{π} e^{- ∣ x ∣} \cos (nx) dx = \frac{2}{π} \int_{0}^{π} e^{- x} \cos (nx) dx$ $\Rightarrow \frac{π}{2} a_{n} = Re (\int_{0}^{π} e^{- x + inx} dx) and \int_{0}^{π} e^{(- 1 + in) x} dx$ $= {[\frac{1}{- 1 + in} e^{(- 1 + in) x}]}_{0}^{π} = \frac{- 1}{1 - in} {e^{(- 1 + in) π} - 1}$ $= - \frac{1 + in}{1 + n^{2}} {{(- 1)}^{n} e^{- π} - 1} = \frac{1 - {(- 1)}^{n} e^{- π}}{1 + n^{2}} + i (. . .) \Rightarrow$ $a_{n} = \frac{2}{π} \times \frac{1 - {(- 1)}^{n} e^{- π}}{1 + n^{2}} \Rightarrow$ $a_{0} = \frac{2}{π} (1 - e^{- π}) \Rightarrow \frac{a_{0}}{2} = \frac{1 - e^{- π}}{π} \Rightarrow$ $e^{- ∣ x ∣} = \frac{1 - e^{- π}}{π} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{1 - {(- 1)}^{n} e^{- π}}{1 + n^{2}} \cos (nx) \Rightarrow$ $x = 0 \Rightarrow 1 = \frac{1 - e^{- π}}{π} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{1}{1 + n^{2}} - \frac{{2 e}^{- π}}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1}$ $\Rightarrow 1 = \frac{1 - e^{- π}}{π} + \frac{2}{π} x - \frac{{2 e}^{- π}}{π} y$ $x = π \Rightarrow e^{- π} = \frac{1 - e^{- π}}{2} + \frac{2}{π} \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{1 - {(- 1)}^{n} e^{- π}}{n^{2} + 1} \Rightarrow$ $e^{- π} = \frac{1 - e^{- π}}{2} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1} - \frac{{2 e}^{- π}}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1}$ $= \frac{1 - e^{- π}}{2} + \frac{2}{π} y - \frac{{2 e}^{- π}}{π} x we get the system$ ${\begin{cases} \frac{2}{π} x - \frac{{2 e}^{- π}}{π} y = 1 - \frac{1 - e^{- π}}{π} \Rightarrow \\ - \frac{{2 e}^{- π}}{π} x + \frac{2}{π} y = e^{- π} - \frac{1 - e^{- π}}{2} \end{cases}$ ${\begin{cases} 2 x - {2 e}^{- π} y = π - 1 + e^{- π} \\ - {2 e}^{- π} x + 2 y = π e^{- π} - \frac{π - π e^{- π}}{2} \end{cases}$ $Δ_{s} = 4 - {4 e}^{- 2 π} \Rightarrow x = \frac{Δ_{x}}{Δ}$ $Δ_{x} = \| \begin{matrix} π - 1 + e^{- π} - {2 e}^{- π} \\ π e^{- π} - \frac{π - π e^{- π}}{2} 2 \end{matrix} \| = . . . .$ $x = \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1} . . . . be continued . . .$
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