I am trying to prove that the power series $$ \sum_{k=1}^{\infty} \frac{x^{k}}{k(k+1)} $$ converges uniformly on $-1 \leq x \leq 1$.
I am planning to use the Weierstrass-M test. This means that if $$ |u_{k}(x)| \leq m_{k} $$ for all $x$ in some interval $I$, with $m_{k} \geq 0$, and if then $$ \sum\limits_{k=1}^{\infty} m_{k} <\infty $$ converges, then the series $$ \sum\limits_{k=1}^{\infty} u_{k} (x) = f(x) $$ converges uniformly.
Now I argue as follows:
We look at the hardest part to show convergence, hence that is at the points where the terms are largest, i.e. at $x=-1$ and $x=1$. Now we take $x = 1$. We then find $$ \left| \frac{x^{k}}{k(k+1)} \right| \leq \left| \frac{1}{k(k+1)} \right| = \left| \frac{1}{k^{2} + k} \right| \leq \left| \frac{1}{k^{2}} \right| = m_{k} , $$ which implies $$ \sum\limits_{k=1}^{\infty} m_{k} = \sum_{k=1}^{\infty} \frac{1}{k^{2}} = \frac{\pi^2}{6}< \infty $$ and we conclude that indeed the power series converges uniformly.
Now my question is: is my reasoning correct? It feels very obvious and I am not sure if what I wrote is valid.
Your reasoning is correct. The exact value of $\sum_{k=1}^\infty m_k$ is not relevant for the M-test, only that the series is convergent.
You could also choose $$ \left| \frac{x^{k}}{k(k+1)} \right| \leq \frac{1}{k(k+1)} = m_k $$ for the comparison, that is a telescoping series: $$ \sum_{k=1}^\infty m_k = \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+1}\right) = 1 $$