Maclaurin series of $\frac{1}{1+x^2}$ I'm stumped here. I''m supposed to find the Maclaurin series of $\frac1{1+x^2}$, but I'm not sure what to do. I know the general idea: find $\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$. What I want to do is find some derivatives and try to see if there's a pattern to their value at $0$. But after the second derivative or so, it becomes pretty difficult to continue. I know this:
$$f(0) = 1$$
$$f'(0) = 0$$
$$f''(0) = -2$$
$$f^{(3)}(0) = 0$$
$$f^{(4)}(0) = 0$$
But when trying to calculate the fifth derivative, I sort of gave up, because it was becoming too unwieldly, and I didn't even know if I was going somewhere with this, not to mention the high probability of making a mistake while differentiating.
Is there a better of way of doing this? Differentiaing many times and then trying to find a pattern doesn't seem to be working.
 A: $1$) Write down the series for $\frac{1}{1-t}$.  You have probably have already seen this one. If not, it can be computed by the method you were using on $\frac{1}{1+x^2}$.  The derivatives are a lot easier to get a handle on than the derivatives of $\frac{1}{1+x^2}$.
$2$) Substitute $-x^2$ for $t$, and simplify.
Comment: It can be quite difficult to find an expression for the $n$-th derivative of a function.  In many cases, we obtain the power series for a function by "recycling" known results.  In particular, we often get new series by adding known ones, or by differentiating or integrating known ones term by term.  Occasionally, substitution is useful.  
A: Since part of your question (yes, I know this is now rather an old question) is related to the matter of finding the higher derivatives for $ \ f(x) \ = \ \frac{1}{1 \ + \ x^2} \ $ at $ \ x=0 \ $ , you might wish to look at what I wrote for computing the derivatives of $ \ \frac{1}{1 \ - \ x^2} \ \ , $  applying the "higher-derivatives" version of the Product Rule and implicit differentiation.  What needs to be adapted to your function is that, starting from  $ \ f(x) · (1+x^2) \ = \ 1 \ \ , $ the second function has derivatives $ \ g'(x) \ = \ 2x  \ \ , \ \  g''(x) \ = \ 2  \ \ , $ and $ \ g^{(k)}(x) \ = \ 0 \ \ , \text{for} \ k \ge 3 \ \ .  $  The pattern for the derivatives becomes
$$ f^{(n)}(0) \ \ = \ \ \left\{  \begin{array}{rcl} 0 \ \ , \ \mbox{for} \ n \ \text{odd}  \\ (-1)^m · n! \ \ , \mbox{for} \ n = 2m \ \text{even} \end{array} \right. \ \ . $$
(Your fourth derivative should equal $ \ 24 \ \ ; $ the sixth equals $ \ -720 \ $ . )
Upon inserting these values into the   Maclaurin terms   $ \ \frac{f^{(2m)}(0)}{(2m)!} \ x^{2m} \ , $ you will have $ \ (-1)^m · x^{2m} \ \ . $ Since $ \ f(x) \ $ is an even function, we expect to arrive at  a series having  only even powers of $ \ x \ $ , and the signs of the  coefficients alternate.
