Define for every $n \geq 2$ the function $$ f_n(x) = \frac{x^n}{n(n-1)}. $$ Aim: show that $\sum_{n=2}^\infty f_n(x) = x + (1-x)\log(1-x)$ for $|x|<1$.
Attempt: Firstly we find the ratio of convergence of this power series. Write $a_n = \frac{1}{n(n-1)}$. Then we have $$ \left| \frac{a_{n+1}}{a_n} \right| =\left|\frac{1}{(n+1)n} \cdot \frac{n(n-1)}{1} \right| = \left( \frac{n-1}{n+1} \right) \to 1 \textrm{ as } n \to \infty. $$ Hence we have have $R=1$.
Moreover, this convergence is uniform. Using the Weierstrass M-test, we have for $n \geq 2$ that $$ \left| \frac{x^n}{n(n-1)} \right| \leq \frac{1}{n(n-1)} =: M_n \textrm{ for all } x \in [0,1]. $$
The task now remains to show that the sum equals the expression given before. Let's first analyse the partial sums:
\begin{align*}
s_n = \sum_{k=2}^n \frac{1}{k(k-1)} x^k = \sum_{k=2}^n \left( \frac{1}{k} - \frac{1}{k-1} \right) x^k = \sum_{k=2}^n \left( \frac{1}{k}x^k - \frac{1}{k-1}x^k \right) = \frac{1}{n}x^n - x^2.
\end{align*}
Hence, we have
$$
\sum_{n=2}^\infty f_n(x) = \lim_{n \to \infty} s_n = -x^2 \lim_{n \to \infty} \frac{1}{n}x^n
$$
Here I am stuck, it seems I'm not getting anywhere close to the desired output, and I'm not sure if what I did so far is correct at all. Any ideas appreciated, thanks.
This seems overly complicated.
$f_n''=x^{n-2}$, so we can integrate $$(\sum_{n=2}^{\infty} f_n (x))''=\sum_{n=2}^{\infty} x^{n-2}=\sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}$$
This holds for all $|x|<1$, as $1$ is the radius of convergence for $\sum f_n''$.
It only remains to integrate it twice to the result.