Partial Sum of this function What are the steps to finding the partial sum formula?
$$\sum\limits_{k=1}^\infty\frac{k}{2^k}=2$$
The professors asked us to search for its partial sum and then prove it by induction
 A: It really is writing down partial sums that you recognize a pattern, which allows you to find the general term of the sequence of partial sums, namely the partial sum formula. As usual, practice is the key.
Let $\displaystyle s_n=\sum_{k=1}^n \frac{k}{2^k}.$ We have $$s_1=\frac{1}{2} \\ s_2 = \frac{1}{2}+\frac{2}{4}=\frac{4}{4} \\ s_3=\frac{4}{4}+\frac{3}{8}=\frac{11}{8} \\ s_4=\frac{11}{8}+\frac{4}{16}=\frac{26}{16} \\ \vdots$$
Do the denominators look familiar? As for the numerators, a technique called finite difference turns out to do the job. That is computing the difference between consecutive terms of a sequence, looking, again, for a pattern. If we denote with $a_n$ the numerator of $s_n$ we have $$b_2=a_2-a_1=3 \\ b_3=a_3-a_2=7 \\ b_4=a_4-a_3=15 \\ \vdots $$ If the denominators looked familiar, I'm sure these do as well. Once you find $b_n$, you're almost done, since $$a_n=1+\sum_{k=2}^nb_k,$$ where $\sum_{k=2}^nb_k$ will be easy to evaluate thanks to the identity $\sum_{k=0}^n2^k=2^{n+1}-1.$
In the end, you'll find what Marco wrote, and the proof is as he showed.
A: Let's prove it using induction. We prove that$$\sum_{k=1}^{N}\frac{k}{2^{k}}=2^{-N}\left(-N+2^{N+1}-2\right).$$
 For $N=1$
  works. Assume is true for $N-1$
 , then$$\sum_{k=1}^{N}\frac{k}{2^{k}}=\sum_{k=1}^{N-1}\frac{k}{2^{k}}+\frac{N}{2^{N}}=2^{1-N}\left(-N+2^{N}-1\right)+N2^{-N}=2^{-N}\left(-N+2^{N+1}-2\right).$$
 Hence$$\sum_{k=1}^{\infty}\frac{k}{2^{k}}=\lim_{N\rightarrow\infty}2^{-N}\left(-N+2^{N+1}-2\right)=2.$$
