Limit of $\sum_{r=1}^{n} \frac{r}{2^r}$ as $n\to \infty$

Question

Consider the sequence $u_{n}=\sum\limits_{r=1}^{n} \frac{r}{2^r}$ with $n \ge 1$. Then the limit of $u_{n}$ as $ n \rightarrow \infty$ is ?

I actually treated each term as an AGP and got $$u_{n}=2-\frac{1}{2^{r+1}}-\frac{n}{2^{r+1}}$$

But how to get the limit?

That formula for $u_n$ can't be right. When $n$ increases, $u_n$ ought to increase. — Christopher Carl Heckman, May 06 '16 at 07:57
You actually want to find $\sum_{r=1}^{\infty}rx^r$ for $x=\frac{1}{2}$. Do you know how to evaluate $\sum_{r=1}^{\infty}rx^r$? — Galc127, May 06 '16 at 08:04
See also: Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? and other posts linked there. — Martin Sleziak, May 29 '19 at 08:07
If you are interested also in the finite sums, there are also several posts that you could check such as: How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$?, How do we get the result of the summation $\sum\limits_{k=1}^n k \cdot 2^k$? or What is the sum of $\sum\limits_{i=1}^{n}ip^i$? — Martin Sleziak, May 29 '19 at 08:10

score 5 · Accepted Answer · answered May 06 '16 at 08:06

This is a well-known result due to Nicole Oresme \begin{align} \sum_{n=1}^\infty \frac{n}{2^n}=2 \end{align} Here is the proof

Consider the function \begin{align*} f(z)=\frac{1}{1-z} \end{align*} The power series expansion of $f(z)$ is \begin{align} f(z)=\sum_{n=1}^{\infty}z^n \end{align} Now take the derivative of $f$, which is \begin{align*} f'(z)=\frac{1}{(1-z)^2}=\sum_{n=1}^{\infty}nz^{n-1} \end{align*} Multiplying by $z$ \begin{align} zf'(z)=\frac{z}{(1-z)^2}=\sum_{n=1}^{\infty}nz^{n}\tag{1} \end{align} Substituting $z=1/2$ in (1) \begin{align} \sum_{n=1}^\infty \frac{n}{2^n}=\frac{\frac{1}{2}}{\biggl(1-\frac{1}{2}\biggl)^2}=2 \end{align}

score 2 · Answer 2 · answered May 06 '16 at 08:26

Let me do this (a bit non-rigorously) without derivatives.

Let $S=\frac12+\frac24+\frac38+\dots+\frac{r}{2^r}+\cdots$

Then

$$\begin{align} S&=\frac12+\frac24+\frac38+\frac4{16}+\cdots \\&=\frac12+\frac12\left(\frac22+\frac34+\frac48+\cdots\right) \\&=\frac12+\frac12\left(\left(\frac12+\frac12\right)+\left(\frac24+\frac14\right)+\left(\frac38+\frac18\right)+\cdots\right) \\&=\frac12+\frac12\left(\left(\frac12+\frac24+\frac38+\cdots\right)+\left(\frac12+\frac14+\frac18+\cdots\right)\right) \\&=\frac12+\frac12(S+1) \end{align}$$

Solving gives $S=2$.

score 1 · Answer 3 · answered May 06 '16 at 08:06

1

Hint:

$$\sum_{r=1}^\infty r a^r = a \frac{d}{da}\sum_{r=1}^\infty a^r$$

answered May 06 '16 at 08:06

RRL

90,707

I think you should also note that it holds for $|a|<1$, not everywhere. – Galc127 May 06 '16 at 08:07
1

Yes - thanks. The hint is to get the ball rolling. – RRL May 06 '16 at 08:10

Limit of $\sum_{r=1}^{n} \frac{r}{2^r}$ as $n\to \infty$

3 Answers3