n-1-1-n2-n-

Question Number 107783 by Dwaipayan Shikari last updated on 12/Aug/20

\underset{n = 1}{\sum^{\infty}} \frac{1}{n 2^{n}}

Commented by Dwaipayan Shikari last updated on 12/Aug/20

\frac{1}{1 . 2^{1}} + \frac{1}{2 . 2^{2}} + \frac{1}{3 . 2^{3}} + \frac{1}{4 . 2^{4}} + \dots . .

= - (- \frac{1}{2} - \frac{1}{2 . 2^{2}} - \frac{1}{3 . 2^{3}} - \dots)

= - l o g (1 - \frac{1}{2})

= l o g (2)