Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 16 down vote accepted

Your sum is equal to

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

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!

share|cite|improve this answer
This is exactly the sort of manipulation that I was looking for. Thank you. – beethree Jul 11 '11 at 22:36
@beethree: Good. You might want to look at the comment by Aryabhata, at the moment hidden at the end. – André Nicolas Jul 11 '11 at 23:03

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 $\sum_{n>=0} n/4^n = 4/9$

share|cite|improve this answer

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.

share|cite|improve this answer

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 ${\displaystyle {{1 \over 4^m} \over 1 - {1 \over 4}} = {4 \over 3}{1 \over 4^m}}$. 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 $${4 \over 3}\times{1 \over 3} = {4 \over 9}$$

share|cite|improve this answer
$\begin{align} (\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \dots ) + \\ (\frac{1}{4^2} + \frac{1}{4^3} + \dots) + \\ (\frac{1}{4^3} + \dots ) + \\ \dots \end{align}$ – Aryabhata Jul 11 '11 at 22:18
yeah, basically :) – Zarrax Jul 11 '11 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 $$ \sum\limits_{n = 1}^\infty {\frac{n}{{4^n }}} = \frac{{1/4}}{{9/16}} = \frac{4}{9}. $$

share|cite|improve this answer
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. – André Nicolas Jul 12 '11 at 0:51
@user6312: Thanks. Hopefully, I'll make more contributions of this kind. – Shai Covo Jul 12 '11 at 1:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.