Does the $\sum_{n=1}^{\infty} 1/(x^2+n^2)$ converge to a function with a continuous derivative? I saw a solution to this question but I have a few questions regarding it. In the solution, it was proven that the original sum converges uniformly in $\mathbb{R}$. Isn't this unnecessary? Doesn't the question only call to check that the sum of $f(n) = \frac{1}{x^2+n^2} $converges point-wise and that the sum $f'(n)$ converges uniformly to some $g(x)$?
In solving the question, the solution showed how the sum of $f'(n)$ converges uniformly, which I had trouble understanding. Can anyone provide some clarity?
Thanks for your time!
 A: Suppose the $f_n$ are continuously differentiable, and $\sum_n f_n'$ converges uniformly on any bounded set to some $g$ (which must be continuous), and suppose $\sum_n f_n(x)$ converges pointwise to some function $f$.
Let $\phi(x) = \int_{x_0}^x g(t) dt$, then uniform convergence gives
$\phi(x) = \sum_n \int_{x_0}^x f_n'(t) dt = \sum_n (f_n(x)-f_n(x_0)) = f(x) -f(x_0)$. Since $g$ is continuous, it follows that $\phi$ is differentiable and
$\phi'(x) = g(x) = f'(x)$. That is, $f$ is differentiable and $f' = g$.  (It also follows that $\sum_n f_n$ converges uniformly to $f$, but that's not relevant here.)
Now let $f_n(x) = {1 \over x^2+n^2}$, then  have $f_n'(x) = - {2 x \over (x^2+n^2)^2 }$. Since $|f_n(x)| \le {1 \over n^2}$, and $|f_n'(x)| \le 2 { |x| \over n^4}$, we see that $f_n(x)$ converges for all $x$ and that $\sum_n f_n'$ converges uniformly to some continuous function $g$ on any bounded set. Then the previous result shows that $\sum_n f_n$ converges to a differentiable function whose derivative is $g$.
