Can anyone please help me solve an infinite series:
$$\sum_{n=0}^{\infty} \frac{n}{2^{(n+1)}}$$
I put it in Wolfram Alpha and got the result that it converges to $1$
I know that the infinite series:
$$\sum_{n=1}^{\infty} \frac{1}{2^{n}}$$
Converges to $1$.
But my series is quite different as it has an additional $n$ term multiplied in, and I can't quite see how to solve it to arrive at result $1$.
Consider the infinite geometric progression $$ \sum_{n=0}^\infty x^n=\frac1{1-x}.\tag1 $$ Differentiating $(1)$ with respect to $x$ yields $$ \sum_{n=0}^\infty nx^{n-1}=\frac1{(1-x)^2}.\tag2 $$ Multiplying $(2)$ by $x^2$ yields $$ \sum_{n=0}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}.\tag3 $$ Setting $x=\dfrac12$ to $(3)$ yields $$ \sum_{n=0}^\infty \frac{n}{2^{n+1}}=\frac{\left(\frac12\right)^2}{\left(1-\frac12\right)^2}=\large\color{blue}{1}. $$
