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

Question Number 95215 by mathmax by abdo last updated on 24/May/20

calculate \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1}

Answered by mathmax by abdo last updated on 25/May/20

let s (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} x^{4 n + 1} with ∣ x ∣ ⩽ 1 and x \neq - 1

s^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n} = \sum_{n = 0}^{\infty} {(- x^{4})}^{n} = \frac{1}{1 + x^{4}} \Rightarrow

s (x) = \int_{0}^{x} \frac{dt}{t^{4} + 1} + c s (0) = 0 = c \Rightarrow s (x) = \int_{0}^{x} \frac{dt}{t^{4} + 1} and

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = \int_{0}^{1} \frac{dt}{t^{4} + 1} = \int_{0}^{1} \frac{\frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt

= \frac{1}{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}} - 1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt = \frac{1}{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt - \frac{1}{2} \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} dt

= \frac{1}{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} dt (\to t - \frac{1}{t} = u) - \frac{1}{2} \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 2} dt (\to t + \frac{1}{t} = v)

= \frac{1}{2} \int_{- \infty}^{0} \frac{du}{u^{2} + 2} - \frac{1}{2} \int_{+ \infty}^{2} \frac{dv}{v^{2} - 2} we have

\int_{- \infty}^{0} \frac{du}{u^{2} + 2} =_{u = \sqrt{2} α} \int_{- \infty}^{0} \frac{\sqrt{2} d α}{2 (1 + α^{2})} = \frac{1}{\sqrt{2}} {[\arctan α]}_{- \infty}^{0}

= \frac{1}{\sqrt{2}} (\frac{π}{2}) = \frac{π}{2 \sqrt{2}} and \int_{\infty}^{2} \frac{dv}{v^{2} - 2} = - \int_{2}^{\infty} \frac{dv}{(v - \sqrt{2}) (v + \sqrt{2})}

= - \frac{1}{2 \sqrt{2}} \int_{2}^{+ \infty} (\frac{1}{v - \sqrt{2}} - \frac{1}{v + \sqrt{2}}) dv = - \frac{1}{2 \sqrt{2}} {[\ln ∣ \frac{v - \sqrt{2}}{v + \sqrt{2}} ∣]}_{2}^{+ \infty}

= - \frac{1}{2 \sqrt{2}} (- \ln (\frac{2 - \sqrt{2}}{2 \sqrt{2}})) = \frac{1}{2 \sqrt{2}} \ln (\frac{1}{\sqrt{2}} - \frac{1}{2}) \Rightarrow

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