# Calculate the sum of the infinite series $\sum_{n=0}^{\infty} \frac{n}{4^n}$

A previous problem had us solving $\sum_{n=0}^{\infty} \frac{1}{4^n}$ which I calculated to be $\frac{4}{3}$ using a bit of mathematical manipulation.

Wonderful. Thank you for all the prompt responses. Could anyone suggest an alternate technique that does not involve differentiation?

$$\frac{1}{4} +\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+ \cdots +\frac{n}{4^n}+\cdots.$$

Call this sum $S$. Now subtract from $S$ the sum $$\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\cdots.$$ If we do it in the obvious way, term by term, we obtain $$\frac{1}{4^2}+\frac{2}{4^3}+\frac{3}{4^4}+ \cdots.$$

Note that this last sum is $(1/4)S$.

Putting things together, and using your computation for $1+1/4+1/4^2+\cdots$ (not quite, we start at $1/4$) we get $$S-\frac{1}{3}=\frac{S}{4}.$$ Solve for $S$. We find that $S=4/9$.

Comment: The calculation is a little sloppy, it assumes that infinite sums can be manipulated much like finite sums. There are theorems about power series that one could use to justify the manipulations.

But (in this case) we do not need such theorems. Let $S_n$ be the sum of the terms up to the term $n/4^n$. More or less the same sort of calculation as the one I did can be used to find an explicit formula for $S_n$. Then we can calculate $\lim_{n\to\infty}S_n$, and get a fully rigorous derivation.

We could use the results of the calculation of $\sum n/4^n$ to tackle $\sum n^2/4^n$, and so on. But the derivatives approach is certainly slicker!

As you have computed $\sum_{n>=0} x^n$ to be $1/(1-x)$, differentiate the series term-wise and multiply by x, which you can do for $x=1/4$ as the series converges. This gives $\sum_{n>=0} n x^{n} = x/(1-x)^{2}$. Substitute $x=1/4$ and observe that $\sum_{n>=0} n/4^n = 4/9$

Let $f(x)=\sum_{n\geq0}x^n$. This defines a function on $(-1,1)$, equal to $\frac1{1-x}$. Using properties of power series, we know that $$\frac x{(1-x)^2}=xf'(x)=\sum_{n\geq1}nx^n$$ for the same values of $x$. Evaluating this equality at $\tfrac14$ sums your series.

Well, you can do some series manipulations... First you can write $$\sum_{n = 1}^{\infty} {n \over 4^n} = \sum_{n = 1}^{\infty}\,\sum_{m = 1}^n {1 \over 4^n}$$ Note that the above summation is over all $(m,n)$ with $m \leq n$. So if you switch the order of summation you obtain $$\sum_{m = 1}^{\infty}\,\sum_{n = m}^{\infty} {1 \over 4^n}$$ The inner sum is a geometric series with initial term ${\displaystyle{1 \over 4^m}}$ and ratio ${\displaystyle{1 \over 4}}$, so it sums to ${{1 \over 4^m} \over 1 - {1 \over 4}} = {4 \over 3}{1 \over 4^m}$. So the overall sum is $${4 \over 3}\sum_{m = 1}^{\infty} {1 \over 4^m}$$ The sum here is a geometric series that sums to ${\displaystyle{1 \over 3}}$, so your final answer is $${4 \over 3}\times{1 \over 3} = {4 \over 9}$$

• \begin{align} (\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \dots ) + \\ (\frac{1}{4^2} + \frac{1}{4^3} + \dots) + \\ (\frac{1}{4^3} + \dots ) + \\ \dots \end{align} Jul 11, 2011 at 22:18
• yeah, basically :) Jul 11, 2011 at 22:19

A simple probabilistic approach:

Let $X$ be a geometric random variable with probability of success $p$, so that $${\rm P}(X=n)=(1-p)^{n-1} p, \;\; n=1,2,3,\ldots.$$ Then the expectation of $X$ is $${\rm E}(X) = \sum\limits_{n = 1}^\infty {n{\rm P}(X = n)} = \sum\limits_{n = 1}^\infty {n(1 - p)^{n - 1} p} = \frac{p}{{1 - p}}\sum\limits_{n = 1}^\infty {n(1 - p)^n } .$$ Thus $$\sum\limits_{n = 1}^\infty {n(1 - p)^n } = \frac{{1 - p}}{p} {\rm E}(X).$$ On the other hand, from $${\rm E}(X) = p \cdot 1 + (1-p)(1+{\rm E}(X)),$$ we get $${\rm E}(X)=\frac{1}{p}.$$ Finally, $$\sum\limits_{n = 1}^\infty {n(1 - p)^n } = \frac{{1 - p}}{p} {\rm E}(X) = \frac{{1 - p}}{p^2}.$$ Letting $p=3/4$ gives $$\sum\limits_{n = 1}^\infty {\frac{n}{{4^n }}} = \frac{{1/4}}{{9/16}} = \frac{4}{9}.$$

• Nice! In imitation of the book "Proofs that Really Count", which is an attractive introduction to bijective proofs, I had thought of posting a question about Mean Proofs. This would ask for examples of results not in probability theory that can be proved by using the mean. This (along with I think a few others of your posts) is a contribution to the mean proofs collection. Jul 12, 2011 at 0:51
• @user6312: Thanks. Hopefully, I'll make more contributions of this kind. Jul 12, 2011 at 1:15
• The validity of this proof actually depends on the convergence of $E(X)$ which of course is not difficult to prove. Sep 20, 2016 at 16:08