Dear Uncle Colin,

I have been given the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{30} + \frac{1}{144} + ...$, which appears to have a general term of $\frac{1}{k! + (k+1)!}$ - but I can't see how to sum that! Any ideas?

- Series Underpin Maths!

Hi, SUM, and thanks for your message! I'm glad you asked, as it gives me a chance to show off some generating functions!

I'm going to take a slightly long way around this, but it's quite neat and worth going into the detail.

### Rewriting the general term

Let's consider the general term: $\frac{1}{k! + (k+1)!}$. That can be rewritten as $\frac{1}{ k! ( k + 2) }$ (since $(k+1)! = k!(k+1)$), so another way to write the sum is $\frac{1}{0! \times 2} + \frac{1}{1! \times 3} + \frac{1}{2! \times 4} + ...$

Those numbers on the bottom remind me of something... oh yes! They remind me of something being integrated.

### Time for some magic

Now consider the function $g(x) =\frac {x^2}{ 0! \times 2 } + \frac{x^3}{ 1! \times 3} + \frac{x^4} {2! \times 4} + ...$, which is equal to the sum when $x = 1$.

Differentiating gives $g'(x) = \frac{x}{0!} + \frac{x^2}{1!} + \frac{x^3}{2!} + ...$, which is the Maclaurin expansion for $xe^x$, so $g'(x) = xe^x$.

Integrating, we find $\int x e^x \dx$ gives $g(x) = (x-1)e^x + C$. We know that $g(0) = 0$ (because it's $x$s all the way down), so $C = 1$.

Therefore, $g(x) = (x-1)e^x + 1$, which makes $g(1)$, the sum of the original series, equal to 1 $\blacksquare$

### A less sophisticated (read, more obvious) method

Another way to write $\frac{1}{k!(k+2)}$ is $\frac{k+1}{(k+2)!}$. So, rather than $\sum_{k=0}^\infty \frac{1}{k!(k+2)}$, we can work with $\sum_{k=0}^{\infty} \frac{k+1}{k+2}!$.

We can change the index to make that nicer: $\sum_{K=2}^{\infty} \frac{K-1}{K!}$, which is $\sum_{K=2}^{\infty} \frac{1}{(K-1)!} - \frac{1}{K!}$.

Expanding this sum gives $\br{\frac{1}{1!} - \frac{1}{2!}} + \br{\frac{1}{2!} - \frac{1}{3!}} + \br{\frac{1}{3!} - \frac{1}{4!}} + ...$ - and the terms pair up nicely!

The sum from 2 to $N$ of this gives $1 - \frac{1}{N!}$, and as $N$ goes to infinity, the sum of the series converges to 1 $\blacksquare$

I like the first way better, though.

Hope that helps!

- Uncle Colin

## 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.