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
 A: 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.

A: 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 $k<m$ and so for each $x\in A$ we have $$\sum_{n=k}^{m} f_n(x) = \sum_{n=k}^{m} \frac{1}{n}g_n(x)=\sum_{n=k}^{m}\frac{1}{n}(h_n(x)-h_{n-1}(x))=$$ $$-\frac 1{k}h_{k-1}(x)+\sum_{n=k}^{m-1}\frac{1}{n(n+1)} h_n(x)+\frac 1{m}h_m(x).$$
