How can I 'prove' the derivative of this function? Consider the function
$$
f: (-1, 1) \rightarrow \mathbb{R}, \hspace{15px} f(x) = \sum_{n=1}^{\infty}(-1)^{n+1} \cdot \frac{x^n}{n}
$$
I am required to show that the derivative of this function is
$$
f'(x) = \frac{1}{1+x}
$$
I have attempted to do this using the elementary definition of a limit, as follows:
$$
f'(x) = \lim_{a \rightarrow x} \left( \frac{f(a)-f(x)}{a-x} \right) = \lim_{a \rightarrow x} \left( \frac{ \sum_{n=1}^{\infty} \left[ (-1)^{n+1} \cdot \frac{a^n}{n} \right]- \sum_{n=1}^{\infty}\left[ (-1)^{n+1} \cdot \frac{x^n}{n} \right]}{a-x} \right) \\
= \lim_{a \rightarrow x} \left( \frac{ \sum_{n=1}^{\infty} \left[ (-1)^{n+1} \cdot \frac{a^n - x^n}{n} \right]}{a-x} \right)
$$
but I am unsure of how to proceed from here (assuming my approach is correct. Can someone help me to show this?

You could observe that this is the taylor expansion for $\ln(x+1)$ and take its first derivative, if you already know about the derivatives for $\ln(x+1)$.
Otherwise, you could note that the sequence of functions converges uniformly in its radius of convergence, and the derivative is a continuous and linear function, so
$\begin{align*}\frac{d}{dx} \lim_{k \to \infty} \sum_{n=1}^{k} (-1)^{n+1} \frac{x^n}{n}&=\lim_{k \to \infty} \sum_{n=1}^{k} (-1)^{n+1} \frac{d}{dx}\frac{x^{n}}{n}\\&=\lim_{k \to \infty}\sum_{n=1}^{k}(-1)^{n-1} n\cdot \frac{x^{n-1}}{n}\\&=\sum_{n=0}^{\infty}(-x)^n \end{align*}$
where the penultimate equality comes from the fact that $(-1)^{n+1}=(-1)^{n-1}$. This has the well known geometric series formula in the interval $(-1,1)$, and is precisely $\frac{1}{1+x}$.
