# Find the value of the series $\sum\limits_{n=1}^ \infty \frac{n}{2^n}$ [duplicate]

Find the value of the series $\sum\limits_{n=1}^ \infty \dfrac{n}{2^n}$

The series on expanding is coming as $\dfrac{1}{2}+\dfrac{2}{2^2}+..$

I tried using the form of $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+..$ and then differentiating it but still it is not coming .What shall I do with this?

• This might help
– user297008
Dec 14, 2015 at 12:36
• Looks like the derivative of a geometric series to me Dec 14, 2015 at 12:37
• See this for other ideas. Dec 14, 2015 at 12:38
• Just differentiate $\frac{1}{2(1-x)}=\frac12\sum x^n$ and set $x=\frac12$. Dec 14, 2015 at 12:39

$$\sum_{n=1}^{\infty}\frac{n}{2^n}=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{n}{2^n}=\lim_{m\to\infty}\frac{-m+2^{m+1}-2}{2^m}=$$ $$\lim_{m\to\infty}\frac{-2^{1-m}+2-2^{-m}m}{1}=\frac{0+2-0}{1}=\frac{2}{1}=2$$