Can I find a power series representation for $\frac {1}{(1+x)^2}$ ONLY by differentiating $\frac {1}{1+x}$?

Question

There are probably much easier to find the power series for $\frac {1}{(1+x)^2}$ than by differentiating $\frac {1}{1/1+x}$, but I think it should be possible. I've gotten extremely close, but my final answer is just a tiny bit different. Here's what I did:

Note that I use the fact that $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$

Firstly, $$\frac{1}{(1+x)^2} = \frac{d}{dx} \frac{1}{1+x} $$

Now, using that, I find the power series representation for $\frac{1}{(1+x)^2}$:

$$\frac{d}{dx} \frac{1}{1+x} = \frac{d}{dx} \frac{1}{1-(-x)} = \frac{d}{dx} \sum_{n=0}^{\infty} (-1)^n(x)^n = \sum_{n=0}^{\infty} \frac{d}{dx} (-1)^n(x)^n = \sum_{n=0}^{\infty} (-1)^nn(x)^{n-1} $$

So my final answer is $$\sum_{n=0}^{\infty} (-1)^nn(x)^{n-1} = \frac{1}{(1+x)^2}$$

But apparently the correct answer is: $$\sum_{n=0}^{\infty} (-1)^n(n+1)(x)^{n} = \frac{1}{(1+x)^2}$$

If anyone could point out my mistake and show me how the latter answer is arrived at I would greatly appreciate it.

Answer 1

Firstly $$\frac{d}{dx}\frac1{1+x}=-\frac1{(1+x)^2}$$ so we have that $$\begin{align} \frac1{(1+x)^2} &=-\sum_{n=0}^{\infty} (-1)^nn(x)^{n-1}\\ &=\sum_{n=0}^{\infty} (-1)^{n+1}n(x)^{n-1}\\ &=\sum_{n=1}^{\infty} (-1)^{n+1}n(x)^{n-1}\\ &=\sum_{n=0}^{\infty} (-1)^{n+2}(n+1)(x)^n\\ &=\sum_{n=0}^{\infty} (-1)^n(n+1)(x)^n\\ \end{align}$$

Answer 2

Just multiply.

Since $\dfrac1{1-x} =\sum_{n=0}^{\infty} x^n $,

$\begin{array}\\ \dfrac1{(1-x)^2} &=(\sum_{n=0}^{\infty} x^n)(\sum_{m=0}^{\infty} x^m)\\ &=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} x^{m+n}\\ &=\sum_{k=0}^{\infty}\sum_{m=0}^{k} x^{k} \qquad k=m+n, n=k-m\\ &=\sum_{k=0}^{\infty}x^k\sum_{m=0}^{k}1\\ &=\sum_{k=0}^{\infty}(k+1)x^k\\ \text{then}\\ \dfrac1{(1+x)^2} &=\sum_{k=0}^{\infty}(k+1)(-1)^kx^k\\ \end{array} $