How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n} $
Solution 1:
Let $S$ be our sum. Then $S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$. Thus $$\begin{align}2S&=1+&\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+\cdots\\ S&=&\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots\end{align} $$ Subtract. We get $$S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots=2.$$
Solution 2:
Hint:
$$ \sum_{k=1}^{\infty} x^k = \frac{x}{1-x} \implies \sum_{k=1}^{\infty} kx^k=\frac{d}{dx} \frac{x}{1-x} . $$