1. Help with Maclaurin series!

Find the Maclaurin series for $\displaystyle [e^(x+1)^2 - 1]/(x+1)^2$ EDIT: the lopsided parenthesis stands for ^( . I don't know why my latex code won't work.

Here's how far I got...

$\displaystyle 1 + (x+1)^2 + [(x+1)^4]/(2!) + [(x+1)^6]/(3!)...$

SUBTRACTED BY...

$\displaystyle 1 - (x^2+2x) + (x^2+2x)^2 - (x^2+2x)^3...$

I'd like to know the summation notion (aka general term) of the series as well if that's possible.

In case you were wondering, I split the expression into two parts with a common denominator and used the maclaurin series of $\displaystyle e^x$ and $\displaystyle 1/(1-x)$ as starting points.

2. Originally Posted by Kaitosan
Find the Maclaurin series for $\displaystyle [e^(x+1)^2 - 1]/(x+1)^2$ EDIT: the lopsided parenthesis stands for ^( . I don't know why my latex code won't work.

Here's how far I got...

$\displaystyle 1 + (x+1)^2 + [(x+1)^4]/(2!) + [(x+1)^6]/(3!)...$

SUBTRACTED BY...

$\displaystyle 1 - (x^2+2x) + (x^2+2x)^2 - (x^2+2x)^3...$

I'd like to know the summation notion (aka general term) of the series as well if that's possible.
$\displaystyle e^{(x+1)^2} - 1 = (x+1)^2 + \frac{(x+1)^4}{2!} + \frac{(x+1)^6}{3!} + ...$

$\displaystyle \frac{e^{(x+1)^2} - 1}{(x+1)^2} = 1 + \frac{(x+1)^2}{2!} + \frac{(x+1)^4}{3!} + ... = \sum_{n=1}^{\infty} \frac{(x+1)^{2(n-1)}}{n!}$

3. Thanks! Often I need someone to point me to the "obvious" lol.