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?

Use the fact that $a^n-x^n=(a-x)\sum_{i=1}^na^{n-i}x^{i-1}$, canceling with the denominator and taking the limit as $a\to x$ we get $(-1)^{n+1}nx^{n-1}$ in the numerator. Cancel the $n$ and now prove that the resulting sum converges to $\frac{1}{1+x}$ using alternating series.
