Let $f:\mathbb{R} \to \mathbb{R}$ be bounded. Let $(g_n)_n$ be the sequence of functions defined by \begin{align} g_n(x) = \sum_{k=2}^n \frac{(-1)^k}{k\ln(k)}f(kx). \end{align} I need to prove that $(g_n)$ converges uniformally on $\mathbb{R}$.
Here is the way I thought of: Since the limit function is not given here and it may be hard to guess, I thought that the best way to prove the uniform convergence is to use Cauchy condition. That is, we begin by $\varepsilon > 0$ and seek $n_0 \in \mathbb{N}$ such that \begin{align} \sup_{x\in \mathbb{R}} |g_n(x) - g_m(x)| = \sup_{x\in \mathbb{R}} \left| \sum_{k=m+1}^n \frac{(-1)^k}{k\ln(k)}f(kx) \right| < \varepsilon \end{align} whenever $n>m>n_0$. My question is how to prove this supremum is less than $\varepsilon$.
I appreciate any help. Thanks in advance.