Sum of the series: $\sum_{n=0}^\infty\frac{1+n}{3^n}$

Question

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.

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. — Adrián Barquero, Dec 16 '10 at 19:01
@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. — Michael Chen, Dec 17 '10 at 06:09
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 :-) — Aryabhata, Dec 17 '10 at 16:31
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 $\sum_{n=0}^\infty (n+1)x^n$ — Martin Sleziak, Jan 09 '17 at 13:53

Adrián Barquero · Accepted Answer · 2010-12-16T18:30:26.570

10

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

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

edited Dec 16 '10 at 18:30

answered Dec 16 '10 at 18:21

Adrián Barquero

15,124

Careful, that's not the series in the question. The sum of the series in the question is 9/4. – Derek Jennings Dec 16 '10 at 18:35
@Derek Oh, yes but I thought the one that Michael could not evaluate was this, the one that sums up to $3/4$. – Adrián Barquero Dec 16 '10 at 18:39
án: Could be, it's not clear. – Derek Jennings Dec 16 '10 at 18:44
Ah, thank you! I forgot about differentiating. – Michael Chen Dec 16 '10 at 18:47
Need to verify that differentiating term by term is valid... – vonbrand Feb 01 '13 at 23:04
@vonbrand Dear vonbrand, it is a standard fact that power series can be differentiated term by term inside their circle or interval of convergence. See here for instance. – Adrián Barquero Feb 02 '13 at 00:40
@AdriánBarquero, forgot about that... too inmersed in formal series lately ;-) – vonbrand Feb 02 '13 at 00:57

SJR · Answer 2 · 2010-12-16T19:25:55.877

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.

The coefficients are stirling numbers: http://math.stackexchange.com/questions/4317/generalizing-sum-n-1-infty-n2-xn-to-sum-n-1-infty-np-xn/4344#4344 — Aryabhata, Dec 16 '10 at 19:15

score 7 · Answer 3 · answered Dec 16 '10 at 19:25

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:

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

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

score 5 · Answer 4 · answered Dec 16 '10 at 18:24

5

Consider for $|x| < 1$

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

Differentiate and set $x=1/3.$

answered Dec 16 '10 at 18:24

Derek Jennings

13,531

score 1 · Answer 5 · answered Jan 09 '17 at 14:57

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

$$\sum_{n=0}^\infty{n+1\over3^n}={9\over4}$$

Sum of the series: $\sum_{n=0}^\infty\frac{1+n}{3^n}$

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