functional series with exponential

Question

I have the functional series $$ f(x)=\sum_{n=0}^{\infty}\frac{e^{-nx}}{n^2+1} $$ I want to find the domain and if the function is continuous in its domain.

My attempt: By root criteria we have that, if $f_n(x)=\frac{e^{-nx}}{n^2+1}$, then $$ \lim_{n\rightarrow\infty}\sqrt[n]{f_n}=e^{-xn} $$ So, the series converges if $x>0$. It is easy to see that this also converges if $x=1$, so, the domain of $f$ is $[0,\infty)$.

Now, we have that $$ \left|\sum_{n=0}^{\infty}\frac{e^{-nx}}{n^2+1}\right|\leq\sum_{n=0}^{\infty}\frac{1}{n^2+1} $$

So, $f$ is absolutely convergent in $[0,\infty)$ and, as each $f_n$ is continuous, $f$ is continuous too.

Is my proof right?

I have trouble with the differentiability part, can someone give me a hint?

Answer 1

You don't really need to use any sophisticated criterion. If $x<0$, you have $$ \lim_{n\to\infty}\frac{e^{-nx}}{1+n^2}=\infty, $$ so the series cannot converge. For $x\geq0$, as you say, $$ \sum_n\frac{e^{-nx}}{1+n^2}\leq\sum_n\frac1{1+n^2}, $$ so the series converges uniformly.

Similarly, the series of the derivatives $$-\sum_n\frac{ne^{-nx}}{1+n^2}$$ converges uniformly on $[r,\infty)$ for any $r>0$. So $f$ is differentiable on $(0,\infty)$. One can check explicitly that the derivative does not exist at $0$.

Answer 2

$$\sum_{n\geq 0}\frac{e^{-nx}}{n^2+1}$$ is an absolutely convergent series of continuous functions (hence a continuous function) for any $x\geq 0$. By denoting such function as $f(x)$ we also have $$ f(x)=\sum_{n\geq 0} e^{-nx} \int_{0}^{+\infty} \sin(s) e^{-ns}\,ds=\int_{0}^{+\infty}\sin(s)\sum_{n\geq 0}e^{-n(s+x)}\,ds=\int_{0}^{+\infty}\frac{\sin(s)}{e^{x+s}-1}\,ds $$ which allows to prove $f(0)=\frac{\pi\coth\pi-1}{2}$.