Pointwise, uniform, normal and $L^p$ convergence of $\sum\limits_{n=1}^{\infty}\frac{e^{-nx}}{1+n^2}$

Question

Study the pointwise, uniform, normal and on $L^p$ convergence of the series

$$\sum_{n=1}^{\infty}\frac{e^{-nx}}{1+n^2}$$

My approach

Let $f_n(x)=\frac{e^{-nx}}{1+n^2}$, it's easy to show that:

1. $f_n(x)\ge 0 \ \ \ \forall x\in\mathbb{R}$;

2. $\lim_{n\to+\infty}f_n(x)=0 \iff x\in [0,+\infty)$

3. $\sum_{n=1}^{+\infty}\frac{e^{-nx}}{1+n^2}\le\sum_{n=1}^{+\infty}\frac{1}{n^2+1}\le\sum_{n=1}^{+\infty}\frac{1}{n^2}<+\infty$

where the first inequality of (3) is true for $x\ge0$. From (3) I know that the series is normally convergent on $[0,+\infty)$, hence it converges pointwise and uniformly on the $[0,+\infty)$.

Now $\sum\frac{e^{-nx}}{n^2+1}$ is a normal convergent series in $[0,+\infty)$ because $\sum\frac{1}{n^2}$ converges.

Let $S_k(x)=\sum_{n=1}^{k}\frac{e^{-nx}}{1+n^2}$ and $S(x)=\sum_{n=1}^{+\infty}\frac{e^{-nx}}{1+n^2}$, how can I prove that

$$\|S-S_k\|_{p}^{p}=\int_{0}^{+\infty}\left|\sum_{n=k+1}^{+\infty}\frac{e^{-nx}}{n^2+1}\right|^pdx \longrightarrow 0$$ for $k\to +\infty$.

My first idea is to use the Lebesgue dominated convergence theorem: if so, the limit operator goes trough the integral, hence $\|S-S_k\|_{p}^{p}\to 0$ for $k\to +\infty$ because $$\sum_{n=k+1}^{+\infty}\frac{e^{-nx}}{n^2+1}$$ is the tail of a convergent series (then $\sum_{n=k+1}^{+\infty}\frac{e^{-nx}}{n^2+1}\to 0$ for $k\to+\infty$).

Now I have many troubles to find $g(x)$ such that $\left|\sum_{n=k+1}^{+\infty}\frac{e^{-nx}}{n^2+1}\right|^p\le g(x)$ and $g(x)\in L^1([0,+\infty))$

I try this:

$$\sum_{n=k+1}^{+\infty}\frac{e^{-nx}}{n^2+1}\le\sum_{n=k+1}^{+\infty}e^{-n x}=\frac{e^{-(k+1)x}}{1-e^{-x}}$$

and now I'm stuck because of $1-e^{-x}$.

Every other inequalities I found are useless to find a proper $g(x)$. Maybe there is a easier way to solve this. Need help. Thanks.

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Edit: This part

where the first inequality of (3) is true for $x\ge0$. From (3) I know that the series is normally convergent on $[0,+\infty)$, hence it converges pointwise and uniformly on the $[0,+\infty)$

is logically incorrect, in fact 3. must be $$3. \ \frac{e^{-nx}}{n^2+1}\le\frac{1}{n^2+1}\le\frac{1}{n^2}$$ and so $\sum\frac{e^{-nx}}{n^2+1}$ is a normal convergent series in $[0,+\infty)$ because $\sum\frac{1}{n^2}$ converges.

Answer 1

$$\left|\sum_{n=k+1}^{+\infty}\frac{e^{-nx}}{n^2+1}\right|^p\le e^{-px} \cdot \left(\sum_{n\geq 1}\frac{1}{(n^2+1)}\right)^p = c^p \cdot e^{-px}$$ And $$ \int_0^{+\infty} c^p \cdot e^{-px} = \frac{c^p}{p} < + \infty $$ So you can use dominated convergence theorem and show the $L^p$ convergence.