October 5, 2006

Find the sum of the series \displaystyle\sum_{n=0}^\infty \frac{(x+2)^n}{(n+3)!}.

Solution. Consider the series

e^x=\sum_{n=0}^\infty \frac{x^n}{n!} = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\cdots

We can make this series look more like our given series by replacing x by x + 2:

e^{x+2}=\sum_{n=0}^\infty \frac{(x+2)^n}{n!} = 1+(x+2)+\frac{(x+2)^2}{2!} + \frac{(x+2)^3}{3!}+\cdots

We can make the given series look like this one if we can make the the number in the denominator whose factorial we're taking the same as the exponent in the numerator.

$\begin{eqnarray}\sum_{n=0}^\infty \frac{(x+2)^n}{(n+3)!}&=& \frac1{(x+2)^3}\sum_{n=0}^\infty \frac{(x+2)^{n+3}}{(n+3)!} \\ &=& (x+2)^{-3}\left[\frac{(x+2)^3}{3!} + \frac{(x+2)^4}{4!}+\cdots \right]\end{eqnarray}$

We see that the series between the brackets is just the series for e^{x+2} with the first three terms missing. So

\sum_{n=0}^\infty \frac{(x+2)^n}{(n+3)!} = (x+2)^{-3}\left[ e^{x+2}-1-(x+2)-\frac{(x+2)^2}{2!}\right]