calculate-n-1-1-n-n-3-n-2n-1-Inspired-from-Sir-Ghaderi-s-post-

Question Number 162424 by mnjuly1970 last updated on 29/Dec/21

c a l c u l a t e

Ω = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} n}{3^{n} (2 n - 1)} = ?

- I n s p i r e d f r o m S i r G {h a d e r i}^{'} s p o s t -

Answered by Ar Brandon last updated on 29/Dec/21

Ω = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} n}{3^{n} (2 n - 1)} = \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(- \frac{1}{3})}^{n} + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(- \frac{1}{3})}^{n} \frac{1}{2 n - 1}

= - \frac{1}{8} + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(- \frac{1}{3})}^{n} \int_{0}^{1} x^{2 n - 2} d x = - \frac{1}{8} + \frac{1}{2} \int_{0}^{1} \frac{1}{x^{2}} \underset{n = 1}{\sum^{\infty}} {(- \frac{x^{2}}{3})}^{n} d x

= - \frac{1}{8} - \frac{1}{2} \int_{0}^{1} \frac{1}{x^{2}} \cdot \frac{\frac{x^{2}}{3}}{1 + \frac{x^{2}}{3}} d x = - \frac{1}{8} - \frac{1}{2} \int_{0}^{1} \frac{1}{x^{2} + 3} d x

= - \frac{1}{8} - {[\frac{1}{2 \sqrt{3}} \arctan (\frac{x}{\sqrt{3}})]}_{0}^{1} = - \frac{1}{8} - \frac{π}{12 \sqrt{3}}

Commented by mnjuly1970 last updated on 29/Dec/21

t h a n k y o u s o m u c h s i r b r a n d o n

Commented by Ar Brandon last updated on 29/Dec/21

My pleasure, Sir!