calculate Σ_(n=0) ^∞ (1/(n^(2 ) +1)) and Σ_(n=0) ^∞ (((−1)^n )/(n^2 +1))


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Question Number 67559 by Abdo msup. last updated on 28/Aug/19

$c a l c u l a t e \sum_{n = 0}^{\infty} \frac{1}{n^{2} + 1} a n d \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1}$
Commented by ~ À ® @ 237 ~ last updated on 28/Aug/19

$l e t n a m e d f (a) = \underset{n = 0}{\sum^{\infty}} \frac{1}{n^{2} + a^{2}} a n d g (a) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2} + a^{2}}$ $f (a) = \frac{1}{2} \underset{- \infty}{\sum^{\infty}} \frac{1}{n^{2} + a^{2}} = - \frac{1}{2} [R e s (A (z), i a) + R e s (A (z), - i a)] w i t h A (z) = \frac{π c o t (π z)}{z^{2} + 1} (t h a t f o r m u l a c o m e s f r o m t h e R e s i d u t h e o r e m . . .$ $R e s (A (z), i a) = \frac{π c o t (i π a)}{2 i a} = - \frac{π c o t h (π a)}{2 a} c a u s e t a n x = - i t h (i x) \Rightarrow c o t (x) = i c o t h (i x)$ $R e s (A (z), - i a) = \frac{π c o t (- i π a)}{- 2 i a} = - \frac{π c o t h (π a)}{2 a}$ $S o f (a) = \frac{π}{2 a} c o t h (π a)$ $g (a) = \frac{1}{2} \underset{- \infty}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2} + a^{2}} = - \frac{1}{2} [R e s (B, i a) + R e s (B, - i a)] w i t h B (z) = \frac{π c s c (π z)}{z^{2} + 1} = \frac{π}{(z^{2} + 1) s i n (π z)}$ $R e s (B, i a) = \frac{π}{2 i a s i n (i π a)} = \frac{π}{2 i a (\frac{e^{- π a} - e^{π a}}{2 i})} = \frac{π}{- 2 a s h (π a)}$ $R e s (B, - i a) = \frac{π}{- 2 i a s i n (- i π a)} = \frac{π}{- 2 a s h (π a)}$ $S o g (a) = \frac{π}{2 a s h (π a)}$
Commented by mathmax by abdo last updated on 28/Aug/19

$t h a n k y o u s i r .$
Commented by mathmax by abdo last updated on 29/Aug/19
$we have proved tbat e^(−∣x∣) =((1−e^(−π) )/π) +(2/π)Σ_(n=1) ^∞ (((1−(−1)^n e^(−π) ))/(1+n^2 ))cos(nx) (developpement at fourier serie) x=0 ⇒1 =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ (1/(n^2 +1)) −(2/π)e^(−π) Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) ⇒π =1−e^(−π) +2 Σ_(n=1) ^∞ (1/(n^2 +1)) −2 e^(−π) Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) x=π ⇒e^(−π) =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ (((1−(−1)^n e^(−π) ))/(n^2 +1))(−1)^n ⇒ e^(−π) =((1−e^(−π) )/π) +(2/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) −(2/π)Σ_(n=1) ^∞ (e^(−π) /(n^2 +1)) ⇒ π e^(−π) =1−e^(−π) +2 Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) −2 e^(−π) Σ_(n=1) ^∞ (1/(n^2 +1)) let x =Σ_(n=1) ^∞ (1/(n^2 +1)) and y =Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) we get π =1−e^(−π) +2x−2e^(−π) y and πe^(−π) =1−e^(−π) +2y −2e^(−π) x ⇒ { ((2x−2e^(−π) y =π−1+e^(−π) )),((−2e^(−π) x +2y =πe^(−π) −1 +e^(−π) )) :} Δ = determinant (((2 −2e^(−π) )),((−2e^(−π) 2)))=4−4 e^(−2π) ≠0 ⇒ Δ_x = determinant (((π−1+e^(−π) −2e^(−π) )),((πe^(−π) −1 +e^(−π) 2)))=2π−2+2e^(−π) +2e^(−π) (πe^(−π) −1+e^(−π) ) =2π −2 +2 e^(−π) +2π e^(−2π) −2e^(−π) +2 e^(−2π) =2π −2 +4π e^(−2π) Δ_y = determinant (((2 π−1+e^(−π) )),((−2e^(−π) πe^(−π) −1+e^(−π) )))=2πe^(−π) −2+2e^(−π) +2e^(−π) (π−1 +e^(−π) )=2π e^(−π) −2 +2 e^(−π) +2π e^(−π) −2e^(−π) +2e^(−2π) =4π e^(−π) −2 +2e^(−2π) ⇒ x =(Δ_x /Δ) =((2π−2 +4π e^(−2π) )/(4−4e^(−2π) )) =((π−1+2π e^(−2π) )/(2−2 e^(−2π) )) =Σ_(n=1) ^∞ (1/(n^2 +1)) y =(Δ_y /Δ) =((4π e^(−π) −2 +2e^(−2π) )/(4−4e^(−2π) )) =((2π e^(−π) −1 +e^(−2π) )/(2−2e^(−2π) )) =Σ_(n=1) ^∞ (((−1)^n )/(n^2 +1)) Σ_(n=0) ^∞ (1/(n^2 +1)) =1+x =1+((π−1+2π e^(−2π) )/(2−2e^(−2π) )) Σ_(n=0) ^∞ (((−1)^n )/(n^2 +1)) =1+y =1+((2πe^(−π) −1 +e^(−2π) )/(2−2e^(−2π) ))$
$w e h a v e p r o v e d t b a t e^{- ∣ x ∣} = \frac{1 - e^{- π}}{π} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{(1 - {(- 1)}^{n} e^{- π})}{1 + n^{2}} c o s (n x)$ $(d e v e l o p p e m e n t a t f o u r i e r s e r i e)$ $x = 0 \Rightarrow 1 = \frac{1 - e^{- π}}{π} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1} - \frac{2}{π} e^{- π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1}$ $\Rightarrow π = 1 - e^{- π} + 2 \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1} - 2 e^{- π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1}$ $x = π \Rightarrow e^{- π} = \frac{1 - e^{- π}}{π} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{(1 - {(- 1)}^{n} e^{- π})}{n^{2} + 1} {(- 1)}^{n} \Rightarrow$ $e^{- π} = \frac{1 - e^{- π}}{π} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1} - \frac{2}{π} \sum_{n = 1}^{\infty} \frac{e^{- π}}{n^{2} + 1} \Rightarrow$ $π e^{- π} = 1 - e^{- π} + 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1} - 2 e^{- π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1}$ $l e t x = \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1} a n d y = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1} w e g e t$ $π = 1 - e^{- π} + 2 x - 2 e^{- π} y a n d π e^{- π} = 1 - e^{- π} + 2 y - 2 e^{- π} x \Rightarrow$ ${\begin{cases} 2 x - 2 e^{- π} y = π - 1 + e^{- π} \\ - 2 e^{- π} x + 2 y = π e^{- π} - 1 + e^{- π} \end{cases}$ $Δ = \| \begin{matrix} 2 - 2 e^{- π} \\ - 2 e^{- π} 2 \end{matrix} \| = 4 - 4 e^{- 2 π} \neq 0 \Rightarrow$ $Δ_{x} = \| \begin{matrix} π - 1 + e^{- π} - 2 e^{- π} \\ π e^{- π} - 1 + e^{- π} 2 \end{matrix} \| = 2 π - 2 + 2 e^{- π} + 2 e^{- π} (π e^{- π} - 1 + e^{- π})$ $= 2 π - 2 + 2 e^{- π} + 2 π e^{- 2 π} - 2 e^{- π} + 2 e^{- 2 π}$ $= 2 π - 2 + 4 π e^{- 2 π}$ $Δ_{y} = \| \begin{matrix} 2 π - 1 + e^{- π} \\ - 2 e^{- π} π e^{- π} - 1 + e^{- π} \end{matrix} \| = 2 π e^{- π} - 2 + 2 e^{- π}$ $+ 2 e^{- π} (π - 1 + e^{- π}) = 2 π e^{- π} - 2 + 2 e^{- π} + 2 π e^{- π} - 2 e^{- π} + 2 e^{- 2 π}$ $= 4 π e^{- π} - 2 + 2 e^{- 2 π} \Rightarrow$ $x = \frac{Δ_{x}}{Δ} = \frac{2 π - 2 + 4 π e^{- 2 π}}{4 - 4 e^{- 2 π}} = \frac{π - 1 + 2 π e^{- 2 π}}{2 - 2 e^{- 2 π}} = \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1}$ $y = \frac{Δ_{y}}{Δ} = \frac{4 π e^{- π} - 2 + 2 e^{- 2 π}}{4 - 4 e^{- 2 π}} = \frac{2 π e^{- π} - 1 + e^{- 2 π}}{2 - 2 e^{- 2 π}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1}$ $\sum_{n = 0}^{\infty} \frac{1}{n^{2} + 1} = 1 + x = 1 + \frac{π - 1 + 2 π e^{- 2 π}}{2 - 2 e^{- 2 π}}$ $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1} = 1 + y = 1 + \frac{2 π e^{- π} - 1 + e^{- 2 π}}{2 - 2 e^{- 2 π}}$
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