Finding a power series for $\frac{1}{(1+x)^2}$

Question

I'm trying to find a power series representation of: $$\frac{1}{(1+x)^2}$$

I know the following:
$$\frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^nx^n$$

My understanding is that if I differentiate the left and right sides of the equation, I'll get (close to) my answer.

Left side: $$\frac{d}{dx}\frac{1}{1+x} = \frac{-1}{(1+x)^2}$$

Right side: $$\frac{d}{dx}\sum_{n=0}^{\infty}(-1)^nx^{n} = \sum_{n=0}^{\infty}(-1)^nnx^{n-1}$$

Since the left side has a $-1$ on top and the original equation doesn't, my understanding is that I need to multiply the right side (the summation) by $-1$ to compensate:

$$-\sum_{n=0}^{\infty}(-1)^nnx^{n-1}$$

When I check the answer to the original equation in my text (and Wolfram Alpha), it gives the power series as:

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

Can anyone help me understand where I've gone astray?

Answer 1

Hint: There is no reason to compensate for anything.

The RHS is \begin{align*} \sum_{n=0}^\infty(-1)^nnx^{n-1}&=\sum_{n=1}^\infty(-1)^nnx^{n-1}\\ &=\sum_{n=0}^\infty(-1)^{n+1}(n+1)x^{n} \end{align*} Equating LHS and RHS gives \begin{align*} \frac{\color{blue}{-1}}{(1+x)^2}=\sum_{n=0}^\infty(-1)^{n\color{blue}{+1}}(n+1)x^{n} \end{align*} and finally \begin{align*} \frac{1}{(1+x)^2}=\sum_{n=0}^\infty(-1)^n(n+1)x^n \end{align*} as it should be.