# Compute the limit $\sum_{n=1}^{\infty} \frac{n}{2^n}$

Compute the limit
$$\sum_{n=1}^{\infty} \frac{n}{2^n}$$

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By ratio test, $\displaystyle \lim_{x\to \infty}\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^{n}}}=\lim_{x\to \infty}\frac{n+1}{2n}=\frac 1 2<1$.

Therefore $\displaystyle\sum_{n=1}^\infty \frac n {2^n}$ convergent.

$\displaystyle S=\sum_{n=1}^\infty \frac n {2^n}=\frac 1 2 + \frac 2 {2^2} + \frac 3 {2^3}+…$ (i)

$2*$(i): $\displaystyle 2S=1 + \frac 2 {2} + \frac 3 {2^2}+…$ (ii)

(ii) $-$ (i): $\displaystyle S=1+\frac 1 2 + \frac 1 {2^2} +…$

Can you take it from here? (Answer: $S=2$)

p.s. This method requires just elementary skills. 🙂

Hint: Consider the power series with those coefficients, for $x=1$. Remember that we can integrate term-by-term within the convergence radius.

Hints:
(1) $f(x)=\frac1{1-x}=\sum_{n=0}^\infty x^n$ converges uniformly for $|x|\leq r$ for any $r<1$ and hence is differentiatable.
(2) $f'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac1x\sum_{n=1}^\infty nx^{n}$