calculate Σ_(n=1) ^∞ (((−1)^n )/(16n^2 −1))


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Question Number 55276 by maxmathsup by imad last updated on 20/Feb/19

$c a l c u l a t e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{16 n^{2} - 1}$
Commented by maxmathsup by imad last updated on 08/Apr/19
$let S =Σ_(n=1) ^∞ (((−1)^n )/(16n^2 −1)) ⇒S =Σ_(n=1) ^∞ (((−1)^n )/((4n−1)(4n+1))) =(1/2)Σ_(n=1) ^∞ (−1)^n { (1/(4n−1)) −(1/(4n+1))} ⇒2S =Σ_(n=1) ^∞ (((−1)^n )/(4n−1)) −Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) we have proved that Σ_(n=0) ^∞ (((−1)^n )/(4n+3)) =(π/(4(√2))) (Q55268) ⇒ Σ_(n=1) ^∞ (((−1)^n )/(4n−1)) =_(n=p+1) Σ_(p=0) ^∞ (((−1)^(p+1) )/(4p+3)) =−Σ_(p=0) ^∞ (((−1)^p )/(4p+3)) =−(π/(4(√2))) let determine Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) let w(x) =Σ_(n=0) ^∞ (−1)^n (x^(4n+1) /(4n+1)) with x∈[−1,1] we have w(1) =Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) =1+Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) ⇒Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) =w(1)−1 we have w^′ (x) =Σ_(n=0) ^∞ (−1)^n x^(4n) =(1/(1+x^4 )) ⇒w(x) =∫_0 ^x (dt/(1+t^4 )) +c c=w(0) =0 ⇒w(x)=∫_0 ^x (dt/(1+t^4 )) and w(1) =∫_0 ^1 (dt/(1+t^4 )) we have ∫_0 ^∞ (dt/(1+t^4 )) =∫_0 ^1 (dt/(1+t^4 )) +∫_1 ^(+∞) (dt/(1+t^4 )) but ∫_1 ^(+∞) (dt/(1+t^4 )) =_(t =(1/x)) −∫_0 ^1 (1/(1+(1/x^4 ))) (−(dx/x^2 )) = ∫_0 ^1 (x^4 /(x^2 (1+x^4 ))) dx =∫_0 ^1 (x^2 /(1+x^4 )) dx =(π/(4(√2))) (result proved) ⇒ ∫_0 ^1 (dt/(1+t^4 )) =∫_0 ^∞ (dt/(1+t^4 )) −(π/(4(√2))) and ∫_0 ^∞ (dt/(1+t^4 )) =_(t =u^(1/4) ) ∫_0 ^∞ (1/(1+u)) (1/4)u^((1/4)−1) du =(1/4) ∫_0 ^∞ (u^((1/4)−1) /(1+u)) du =(1/4) (π/(sin((π/4)))) =(π/(4(1/(√2)))) =((π(√2))/4) ⇒∫_0 ^1 (dt/(1+t^4 )) =((π(√2))/4) −(π/(4(√2))) =((π(√2))/4) −((π(√2))/8) =((π(√2))/8) ⇒Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) =((π(√2))/8) −1 ⇒ S =(1/2){−(π/(4(√2))) −((π(√2))/8) +1} =(1/2){1−((π(√2))/4)} .$
$l e t S = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{16 n^{2} - 1} \Rightarrow S = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(4 n - 1) (4 n + 1)}$ $= \frac{1}{2} \sum_{n = 1}^{\infty} {(- 1)}^{n} {\frac{1}{4 n - 1} - \frac{1}{4 n + 1}} \Rightarrow 2 S = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n - 1} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1}$ $w e h a v e p r o v e d t h a t \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 3} = \frac{π}{4 \sqrt{2}} (Q 55268) \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n - 1} =_{n = p + 1} \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p + 1}}{4 p + 3} = - \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{4 p + 3} = - \frac{π}{4 \sqrt{2}}$ $l e t d e t e r m i n e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} l e t w (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{4 n + 1}}{4 n + 1} w i t h x \in [- 1, 1]$ $w e h a v e w (1) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = 1 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = w (1) - 1$ $w e h a v e w^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n} = \frac{1}{1 + x^{4}} \Rightarrow w (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} + c$ $c = w (0) = 0 \Rightarrow w (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} a n d w (1) = \int_{0}^{1} \frac{d t}{1 + t^{4}}$ $w e h a v e \int_{0}^{\infty} \frac{d t}{1 + t^{4}} = \int_{0}^{1} \frac{d t}{1 + t^{4}} + \int_{1}^{+ \infty} \frac{d t}{1 + t^{4}} b u t \int_{1}^{+ \infty} \frac{d t}{1 + t^{4}} =_{t = \frac{1}{x}} - \int_{0}^{1} \frac{1}{1 + \frac{1}{x^{4}}} (- \frac{d x}{x^{2}})$ $= \int_{0}^{1} \frac{x^{4}}{x^{2} (1 + x^{4})} d x = \int_{0}^{1} \frac{x^{2}}{1 + x^{4}} d x = \frac{π}{4 \sqrt{2}} (r e s u l t p r o v e d) \Rightarrow$ $\int_{0}^{1} \frac{d t}{1 + t^{4}} = \int_{0}^{\infty} \frac{d t}{1 + t^{4}} - \frac{π}{4 \sqrt{2}} a n d \int_{0}^{\infty} \frac{d t}{1 + t^{4}} =_{t = u^{\frac{1}{4}}} \int_{0}^{\infty} \frac{1}{1 + u} \frac{1}{4} u^{\frac{1}{4} - 1} d u$ $= \frac{1}{4} \int_{0}^{\infty} \frac{u^{\frac{1}{4} - 1}}{1 + u} d u = \frac{1}{4} \frac{π}{s i n (\frac{π}{4})} = \frac{π}{4 \frac{1}{\sqrt{2}}} = \frac{π \sqrt{2}}{4} \Rightarrow \int_{0}^{1} \frac{d t}{1 + t^{4}} = \frac{π \sqrt{2}}{4} - \frac{π}{4 \sqrt{2}}$ $= \frac{π \sqrt{2}}{4} - \frac{π \sqrt{2}}{8} = \frac{π \sqrt{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = \frac{π \sqrt{2}}{8} - 1 \Rightarrow$ $S = \frac{1}{2} {- \frac{π}{4 \sqrt{2}} - \frac{π \sqrt{2}}{8} + 1} = \frac{1}{2} {1 - \frac{π \sqrt{2}}{4}} .$
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