I am a calculus TA a bit stumped at a question a student has posed me.

What is the method for finding the sum of the series $$\sum_{n=0}^\infty\frac{1+n}{3^n}$$ If you split the numerator up, one becomes a geometric series summing to 3/2, and the other sums to 3/4 by Wolfram Alpha, but I can't explain why.

  • 2
    $\begingroup$ If the second part of your question is not relevant then I think it would be best if you just edit it and delete that portion of the question, instead of having a message telling people to ignore that part. $\endgroup$ Dec 16, 2010 at 19:01
  • $\begingroup$ @Adrián Barquero - Acknowledged. I spent so much time formatting it to look just right that I didn't have the heart to delete it myself. I will do as you suggest next time. $\endgroup$ Dec 17, 2010 at 6:09
  • $\begingroup$ If you don't want to delete it, you can change your question a bit. For instance, you could have asked what the value of the limit is... Sorry, I deleted it for you :-) $\endgroup$
    – Aryabhata
    Dec 17, 2010 at 16:31
  • $\begingroup$ See also: Limit of this series: $\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$? and the more general question How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ $\endgroup$ Jan 9, 2017 at 13:53

5 Answers 5


Using the usual geometric series, you have

$$\frac{x}{1-x} = \sum_{n=1}^{\infty} x^n $$ thus differentiating you get

$$\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$$ thus

$$\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^2}$$ Then

$$\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^2}$$

So your sum is just

$$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{1/3}{(1-1/3)^2} = \frac{3}{4}$$

  • $\begingroup$ Careful, that's not the series in the question. The sum of the series in the question is 9/4. $\endgroup$ Dec 16, 2010 at 18:35
  • $\begingroup$ @Derek Oh, yes but I thought the one that Michael could not evaluate was this, the one that sums up to $3/4$. $\endgroup$ Dec 16, 2010 at 18:39
  • $\begingroup$ án: Could be, it's not clear. $\endgroup$ Dec 16, 2010 at 18:44
  • $\begingroup$ Ah, thank you! I forgot about differentiating. $\endgroup$ Dec 16, 2010 at 18:47
  • $\begingroup$ Need to verify that differentiating term by term is valid... $\endgroup$
    – vonbrand
    Feb 1, 2013 at 23:04

Michael, there is a general trick here that can be very useful.

Think of $xD$ as an operator on differentiable functions $p(x)$ that "differentiates and then multiplies by $x$". Then polynomials in $xD$ are also operators on differentiable functions, if one interprets powers of $xD$ by composition. For example if $f(x)=2x^{2}+3$ then $f(xD)$ applied to a differentiable function p(x) gives $2xD(xD(p(x)))+3p(x)$. (Think of 3 as the operator "multiplication by 3". )

Now you should verify the following very useful fact:

The operator $p(xD)$ applied term by term to the power series $\sum_{n}a_{n}x^{n}$ gives the power series $\sum_{n}p(n)a_{n}x^{n}$.

In particular, your series has the form $\sum f(n)x^{n}$ evaluated at $x=1/3$, where $f(n)=n+1$. So to compute the sum, apply $1+xD$ to the function $1/(1-x)=1+x+x^{2}+\ldots$ and then evaluate at $x=1/3$.

I think I learned this from Polya and Szego's "Problems and Theorems in Analysis", which is absolutely stuffed with mathematical wisdom.


I believe the other answers about differentiating and such are much cleaner, but I thought I would add that there is an even more elementary way of doing the sum:

$$ S = \sum_{k=1}^{\infty} k b^{-k} $$

by using the standard trick of multiplying by $b^{-1}$ then subtracting:

$$ b^{-1} S = \sum_{k=2}^{\infty} (k-1) b^{-k} $$

$$ S - b^{-1} S = S (1 - b^{-1}) = \sum_{k=1}^{\infty} b^{-k} $$


Consider for $|x| < 1$

$$\sum_{n=1}^\infty x^n = \frac{x}{1-x} .$$

Differentiate and set $x=1/3.$


This is just a different take on (much) earlier answers, mostly to emphasize what can sometimes be gained by playing with the index of summation.

Shifting the index of summation and noting that including $nx^{n-1}$ for $n=0$ adds nothing to the sum, one obtains (for $|x|\lt1$)

$$\sum_{n=0}^\infty(n+1)x^n=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty nx^{n-1}=\left(\sum_{n=0}^\infty x^n\right)'=\left(1\over1-x\right)'={1\over(1-x)^2}$$

Setting $x={1\over3}$ gives



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