Prove that $f(x)=\sum_\limits{n=1}^{\infty} \frac{1}{x^2+n^2}$ is differentiable on $\mathbb{R}$ Prove that $f(x)=\sum_\limits{n=1}^{\infty} \frac{1}{x^2+n^2}$ is differentiable on $\mathbb{R}$.

Solution:
Let $I=(-a,a)$. Each partial sum $f_k(x)=\sum_1^k \frac{1}{x^2+k^2}$ is differentiable on $I$. Also, $f(0)$ is convergent. Since for $x\in I$ 
$$|\frac{-2x}{(x^2+k^2)^2}| \le \frac{2a}{(x^2+k^2)^2}$$ and $\sum_1^{\infty}\frac{2a}{(x^2+k^2)^2}$ is convergent, we have that $f'$ exists on $I$ and $f'_k$ is uniformly convergent to $f'$ on $I$. Since $a$ is arbitrarly large, the claim follows. Is this correct?
 A: Your approach looks fine. The theorem you need to justify your claims is the following (see for example Theorem 2 in this note):

*

*Term by term differentiation theorem: If $\sum f_k(x_0)$ converges at some point $x_0\in [a,b]$ and the series of derivatives $g(x) = \sum f_k'(x)$ converges uniformly on $[a,b]$ then the series $f(x) = \sum f_k(x)$ converges uniformly for all $x\in[a,b]$ and $f(x)$ is differentiable with $f'(x) =g(x)$.

Below is two alternative derivations.

From the definition of the derivative we have
$$f'(x) = \lim_{h\to 0}\frac{f(x+h) -f(x)}{h} = -\lim_{h\to 0}\sum_{n=1}^\infty\frac{2x+h}{((x+h)^2+n^2)(x^2+n^2)}$$
The summand satisfies
$$\left|\frac{2x+h}{((x+h)^2+n^2)(x^2+n^2)}\right| \leq \frac{2|x|+1}{n^4}$$
for $h<1$ so the series above converges uniformly by Weierstrass M-test and we can therefore move the limit $h\to 0$ inside the summation to get
$$f'(x) = -\sum_{n=1}^\infty\lim_{h\to 0}\frac{2x+h}{((x+h)^2+n^2)(x^2+n^2)} = -\sum_{n=1}^\infty \frac{2x}{(x^2+n^2)^2}$$
Since the series above converges for all $x\in\mathbb{R}$ it follows that $f(x)$ is differentiable on $\mathbb{R}$.

Another way to show it is by evaluating the series in closed form
$$f(x) = \left\{\matrix{\dfrac{\pi  \coth (\pi  x)}{2 x}-\dfrac{1}{2 x^2} & x\not =0\\\dfrac{\pi^2}{6} & x=0}\right.$$
from which it's not hard to see, apart from $x=0$ which requires some work, that the resulting function is differentiable on $\mathbb{R}$.
