calculate-n-1-1-n-16n-2-1-

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

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}} .