I want to show that
$$\sum_{n=1}^\infty f_n =\sum_{n=1}^\infty \log\left(1+\frac{x}{n^2}\right)$$
converges on $X=[0, 1]$
My approach was
$$\log\left(1+\frac{x}{n^2}\right) \le \log\left(1+\frac{1}{n^2}\right) \le \log2$$
so
$$\Vert f_n\Vert_{\infty,X} \le \log2 \lt \infty$$
and we know that
$$\lim_{n\to \infty}\frac{1}{n^2} \to 0$$
so
$$\lim_{n\to \infty} \log\left(1+\frac{1}{n^2}\right) \to \log1=0$$
and lets consider $S_i$ the partial sum from $n=1$ to $i$
we have $\forall \epsilon \gt 0, \exists N \in \Bbb{N}$ such that $\forall n,m \gt N$ and $(m \gt n)$
$$\Vert S_m - S_n\Vert_{\infty,X} \lt \epsilon$$
So the serie is Cauchy and we can conclude with Weierstrass criterion that it is uniformly convergent on $X$.
I am just wondering if I made a mistake since I just begun in analysis and I'm never sure.