Summing with Generating Functions

A nice challenge puzzle via Reddit:

Find $\sum_{n=1}^{\infty} \frac{n2^n}{(n+2)!}$

There was a video attached to it that I didn’t watch, something about telescoping sums, but the moment I saw this, I thought: generating functions!

Why would I think such a thing?

The thing that jumped out at me was the $n$ on top: that looks to me like its something I could integrate and divide away. It looks like an infinite sum with factorials on the bottom, which will be related to $e^{something}$. Let’s have a look, shall we?

I’m going to call the whole thing $s$, and write it as $s(x) = 2 \sum_{n=1}^{\infty} \frac{n x^{n-1}}{(n+2)!}$, and we’re eventually going to evaluate it at $x=2$.

But first, I’ll integrate term by term to get $S = \int s(x) \dx = 2\sum_{n=1}^{\infty} \frac{ x^n }{(n+2)!}$, or $C+\frac{x}{3!} + \frac{x^2}{4!} + \dots$. (I’m going to ignore the constant, reasoning that it’ll differentiate away in good time.)

So what is this function? Well, $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$, and I’m mainly interested in matching up the bottoms. So, if I want to something starting with $\frac{x}{3!}$, I’m going to need to subtract the first three terms and divide by $x^2$ to get $S(x) = \frac{2}{x^2}\left( e^x - \frac{1}{2}x^2 - x - 1\right)$.

Now to differentiate, using product rule, to get $s(x) = \frac{2}{x^2}\left(e^x - x - 1\right) - \frac{4}{x^3}\left(e^x - \frac{1}{2}x^2 - x - 1\right)$.

We can evaluate that at $x=2$ to get $\frac{2}{4}\left(e^2 - 3\right) - \frac{4}{8}\left(e^2 - 5\right)$… which is $-\frac{3}{2} + \frac{5}{2}$, which is 1!

Reassuringly, that’s what the video got, too.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

Share

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter