# Uniform Convergence of a Dampened Series of Functions

Let $$\sum_{n=1}^{\infty} g_n(x)$$ converge uniformly to a function $$g(x)$$ on $$A \subseteq \mathbb{R}$$. Can we say that the series defined by $$\sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty} \frac{1}{n}g_n(x)$$ converges uniformly to a function $$f(x)$$ on $$A \subseteq \mathbb{R}$$? If so, prove it. If not, give a counter example.

I have not been able to find a counterexample to this proposition, and my intuition says that this statement is true. When trying to prove it, I attempt to make use of the Cauchy Criterion for the Uniform Convergence of a Series which states that for any $$\epsilon > 0,$$ there exists an $$N \in \mathbb{N}$$ such that $$|g_{m+1}(x)+g_{m+2}(x)+...+g_n(x)| < \epsilon$$ whenever $$n>m \geq N$$ and $$x \in A.$$

Now, I consider the analogous sum for the second series; namely, $$|\frac{1}{m+1}g_{m+1}(x)+\frac{1}{m+2}g_{m+2}(x)+...+\frac{1}{n}g_n(x)|.$$ If we add the restriction that $$g_n(x) \geq 0$$ for all $$n \in \mathbb{N}$$, the result falls out immediately as this sum will be less than or equal to the sum in the preceding paragraph. However, I do not know how to do it for general $$g_n(x).$$

Any help would be appreciated, but please try to only give hints.

Thanks

If $$g$$ is bounded, the answer is yes.
For any $$N \ge 1$$, $$\sum_{n=1}^N \frac{g_n(x)}{n} = \frac{\sum_{n=1}^N g_n(x)}{N}+\int_1^N \frac{\sum_{n \le t} g_n(x)}{t^2}dt$$, so
$$\left|\sum_{n=1}^N \frac{g_n(x)}{n}-\int_1^\infty \frac{\sum_{n \le t} g_n(x)}{t^2}dt\right| \le \frac{\left|\sum_{n=1}^N g_n(x)\right|}{N}+\int_N^\infty \frac{\left|\sum_{n \le t} g_n(x)\right|}{t^2}dt$$. Take $$\epsilon > 0$$ and then $$N_0$$ so that $$|\sum_{n=1}^N g_n(x) - g(x)| \le \epsilon$$ for all $$x \in A$$ and $$N \ge N_0$$. Then for any $$x \in A$$ and any $$N \ge N_0$$, $$\left|\sum_{n=1}^N \frac{g_n(x)}{n}-\int_1^\infty \frac{\sum_{n \le t} g_n(x)}{t^2}dt\right| \le \frac{g(x)+\epsilon}{N}+\int_N^\infty \frac{g(x)+\epsilon}{t^2}dt \le 2\frac{g(x)+\epsilon}{N} \le 2\frac{C+\epsilon}{N_0}$$ is arbitrarily small.
Yes, we can say that. We can even prove that using the following idea. Put $$h_0(x)=0$$ for each $$x\in A$$ and $$h_k=\sum_{n=1}^k g_n$$ for each natural $$k$$. Then $$g_n=h_n-h_{n-1}$$ for each natural $$k and so for each $$x\in A$$ we have $$\sum_{n=k}^{m} f_n(x) = \sum_{n=k}^{m} \frac{1}{n}g_n(x)=\sum_{n=k}^{m}\frac{1}{n}(h_n(x)-h_{n-1}(x))=$$ $$-\frac 1{k}h_{k-1}(x)+\sum_{n=k}^{m-1}\frac{1}{n(n+1)} h_n(x)+\frac 1{m}h_m(x).$$