$(a)$ Starting with the geometric series $\displaystyle \sum_{n=0}^{\infty} x^n$, find the sum of the series $\displaystyle \sum_{n=1}^{\infty} nx^{n-1}\;, \vert x\vert<1$.
$(b)$ Find the sum of the series $\displaystyle \sum_{n=1}^{\infty} nx^n\;, \vert x\vert<1$ and $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n}$.
Here's the image of the question. I know that the sum from $0$ to $\infty$ of part $A$ is the same as the sum to $\infty$ from $1$ if you decrease the power by $1$. So I'm guessing the series will converge, but I don't know how to find the sum ( because $x$ can be both $+ve$ and $-ve$ ( I assumed)) .
For part $A$ I got the sum as $1+x+x^2+x^3$ and so on.
So the next bit will be $1+2x+3x^2+4x^3+\dots +(\infty+1)x^{\infty}$.
My questions is am I doing it right because it seems way too easy. And if not, then what should I do?
Best Answer
There is a closed formula (not written as an infinite sum) for the sum of a geometric series. Find out what that is. Now differentiate series and its sum formula. For part (b) you are just playing with your series from part (a).