1

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

Peter Woolfitt
  • 21,114
BioShock
  • 261
  • http://math.stackexchange.com/questions/647587/sum-of-a-power-series-n-xn OR http://math.stackexchange.com/questions/333192/solve-sum-nxn – lab bhattacharjee Mar 23 '15 at 16:48
  • By partial sum, do you mean that you want a closed form for $$f(k) = \sum_{k=1}^{\color{red}n} \frac{k}{2^k}$$ ...? – DanielV Mar 23 '15 at 16:54
  • Yes! That's what i think, but not sure! He just said find a partial sum of this and prove it.. – BioShock Mar 23 '15 at 16:56

2 Answers2

2

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.

Vincenzo Oliva
  • 7,249
0

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$$\sum_{k=1}^{\infty}\frac{k}{2^{k}}=\lim_{N\rightarrow\infty}2^{-N}\left(-N+2^{N+1}-2\right)=2.$$

Marco Cantarini
  • 33,062
  • 2
  • 47
  • 93