I needed to solve a recurrence relation and used the recursion tree method. I simplified the series that the tree gave me and eventually got to a point where I needed to evaluate $\sum_{k = 0}^{\infty} \frac{-k}{2^k}$. I'm not proficient with series yet, and I don't know how to evaluate this series. I tried arriving at something that looks like a geometric series but to no avail. How to evaluate this series? Thanks!
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Does this answer your question? (In the case of $p=1/2$ specifically.) Proof for Mean of Geometric Distribution See also the answers here: Deriving the mean of the Geometric Distribution – Semiclassical Feb 07 '23 at 15:22
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Yes, thanks!!!! – sag0li Feb 07 '23 at 15:23
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more generally since $(x^n)' = nx^{n-1}$ you can obtain sums $\sum p(k) x^k$ where $p$is polynomial by successive derivations or integrations. Convergence is not an issue with geometric series. I.e. here $\sum \frac {-k}{2^k}=-\frac 12\sum k\left(\frac 12\right)^{k-1}$ – zwim Feb 07 '23 at 15:24