Taylor Expansion for $f(x) = \frac{x}{(1+x^{2})^{2}}$ I found the expansions for 
$$f(x) = \frac{1}{1+x^{2}} = \sum_{k=0}^{\infty}(-1)^{k}x^{2k}$$
$$g(x) = \frac{1}{(1+x^{2})^{2}} = \sum_{k=0}^{\infty}(-1)^{k}x^{4k}$$
$$h(x) = \frac{x}{1+x^{2}} = \sum_{k = 0}^{\infty}(-1)^{k}x^{2k+1}$$
But what would be the proper approach of finding an expansion for 
$$w(x) = \frac{x}{(1+x^{2})^{2}}$$
I think I can either go with:
(1) x* $\frac{1}{(1+x^{2})^{2}}$ which would give me $\sum_{k = 0}^{\infty} (-1)^{k} x ^ {4k + 1}$
or (2) squaring the denominator of $\frac{x}{1 + x^{2}}$ which would give me $\sum_{k = 0}^{\infty} (-1)^{k} x ^ {4k + 2}$.
I think that approach (1) is more correct because it deals with just multiplying by x rather than squaring (1+$x^{2}$) but am I right? Is (1) correct and is my reasoning for its correctness correct? 
 A: Start from
$$\frac1{1+x}=\sum_{k=0}^\infty(-x)^k.$$
Differentiating both members,
$$-\frac1{(1+x)^2}=\sum_{k=1}^\infty k(-x)^{k-1}=-\sum_{k=0}^\infty (k+1)(-x)^k.$$
Now you can substitute $x^2$ for $x$,
$$\frac1{(1+x^2)^2}=\sum_{k=0}^\infty (k+1)(-x^2)^k$$
and multiply by $x$ (your approach (1)),
$$\frac x{(1+x^2)^2}=\sum_{k=0}^\infty (k+1)(-x^2)^kx=\sum_{k=0}^\infty (k+1)(-1)^kx^{2k+1}.$$

To "square the denominator" as in your approach (2), you should perform the product of two series, which is not straightforward.

Alternatively to the "derivative trick" (which needs to be justified by convergence analysis), you can obtain the second series as follows: let
$$h(x)=(1+x)^{-1}\to h(0)=1,\\
h'(x)=(-1)(1+x)^{-2}\to h'(0)=-1,\\
h''(x)=(-1)(1+x)^{-3}\to h''(0)=(-1)(-2)=2!,\\
h'''(x)=(-1)(1+x)^{-4}\to h'''(0)=(-1)(-2)(-3)=-3!,\\\cdots
$$
This obviously gives you the development
$$h(x)=\frac1{1+x}=\sum_{k=0}^\infty\frac{(-1)^kk!}{k!}x^k=\sum_{k=0}^\infty(-x)^k.$$
But for the same "price", you obtain the development of $i(x):=h'(x)$ by shifting the coefficients by one position ($i(0)=h'(0), i'(0)=h''(0), i''(0)=h'''(0),\cdots$):
$$i(x)=h'(x)=-\frac1{(1+x)^2}=\sum_{k=0}^\infty\frac{(-1)^{k+1}(k+1)!}{k!}x^k=-\sum_{k=0}^\infty (k+1)(-x)^k.$$
A: I think the proper approach is as follows. Note that:
$$\frac{d}{dx}\left( \frac{-1}{2(1+x^2)} \right)= \frac{x}{(1+x^2)^2}.$$
The power series for $$\frac{1}{1+x^2}=\sum_{n=0}^{\infty} (-1)^n x^{2n},$$ as you found. Hence differentiating the power series and then multiplying by the $-\frac{1}{2}$ gives the desired result for your function in consideration.
If I were you, I wouldn't try to square a series. Trying to square series usually complicates more things. For example, I refer to a problem in analysis, namely the Basel Problem, and a particular solution, which does something like you are trying to do. Can the Basel problem be solved by Leibniz today?
A: Your expansion of $g (x) $ doesn't seem correct.
try this
$$\frac {1}{(1+x^2)^2}=\frac {1}{1-(-x^4-2x^2)}$$
$$=1+t+t^2+t^3+.... $$
with
$$t^n=(-1)^n (x^4+2x^2)^n $$
$$=(-1)^nx^{2n}\sum_{k=0}^n \binom {n}{k}x^{2k}2^{n-k} $$
