calculate-n-0-1-n-4n-3-

Question Number 55268 by maxmathsup by imad last updated on 20/Feb/19

c a l c u l a t e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 3}

Commented by maxmathsup by imad last updated on 21/Feb/19

l e t S (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 3} x^{4 n + 3} w i t h ∣ x ∣< 1 w e h a v e

\frac{d S}{d x} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n + 2} = x^{2} \sum_{n = 0}^{\infty} {(- x^{4})}^{n} = \frac{x^{2}}{1 + x^{4}} \Rightarrow

S (x) = \int_{0}^{x} \frac{t^{2}}{1 + t^{4}} d t + c b u t c = S (0) = 0 \Rightarrow S (x) = \int_{0}^{x} \frac{t^{2}}{t^{4} + 1} d t a n d

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 3} = {l i m}_{x \to 1} S (x) = \int_{0}^{1} \frac{t^{2}}{1 + t^{4}} d t = I

w e h a v e \int_{0}^{\infty} \frac{t^{2}}{t^{4} + 1} d t = \int_{0}^{1} \frac{t^{2}}{t^{4} + 1} d t + \int_{1}^{+ \infty} \frac{t^{2}}{t^{4} + 1} d t a n d

\int_{1}^{+ \infty} \frac{t^{2}}{t^{4} + 1} d t =_{t = \frac{1}{u}} - \int_{0}^{1} \frac{1}{u^{2} (\frac{1}{u^{4}} + 1)} \frac{- d u}{u^{2}} = \int_{0}^{1} \frac{d u}{1 + u^{4}} \Rightarrow

2 \int_{0}^{1} \frac{t^{2}}{t^{4} + 1} d t = \int_{0}^{\infty} \frac{t^{2}}{t^{4} + 1} d t \Rightarrow \int_{0}^{1} \frac{t^{2}}{t^{4} + 1} d t = \frac{1}{2} \int_{0}^{\infty} \frac{t^{2}}{t^{4} + 1} d t c h a n g e m e n t

t = α^{\frac{1}{4}} g i v e \int_{0}^{\infty} \frac{t^{2}}{t^{4} + 1} d t = \int_{0}^{\infty} \frac{α^{\frac{1}{2}}}{1 + α} \frac{1}{4} α^{\frac{1}{4} - 1} d α = \frac{1}{4} \int_{0}^{\infty} \frac{α^{\frac{3}{4} - 1}}{1 + α} d α

= \frac{1}{4} \frac{π}{s i n (\frac{3 π}{4})} = \frac{π}{4 s i n (\frac{π}{4})} = \frac{π}{4 \frac{\sqrt{2}}{2}} = \frac{π}{2 \sqrt{2}} \Rightarrow \int_{0}^{1} \frac{t^{2}}{t^{4} + 1} d t = \frac{π}{4 \sqrt{2}} \Rightarrow

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 3} = \frac{π}{4 \sqrt{2}} .