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

Question

(1) Let $r$ be a positive real number. Show $\sum_{n=1}^{\infty}\frac{e^{-nx}}{n+1}$ is uniformly convergent on $x\in [r,\infty)$.

(2) Find the value of $\lim_{r\to+0} \int_{r}^{1/r}\sum_{n=1}^{\infty}\frac{e^{-nx}}{n+1}dx$

My approach

(1) Let $f_n(x) = \frac{e^{-nx}}{n+1}$. $f_{n}^{'}(x) < 0$ on the interval, then $\sum_{n=1}^{\infty}f_n(x) < \sum_{n=1}^{\infty}f_n(r) $. For sufficiently large $n$, $e^{-nx} < 1/(n+1)$. So there exists $N$ and $\sum_{n=1}^{\infty}f_n(r) <\sum_{n=1}^{N-1}f_n(r) + \sum_{n=N}^{\infty}\frac{1}{(n+1)^2} < \infty$. From Riemann zeta function and Weierstrass M test, we get the result.

(2) We can assume $r<1$.

$\lim_{r\to+0} \int_{r}^{1/r}\sum_{n=1}^{\infty}\frac{e^{-nx}}{n+1}dx = \lim_{r\to+0} \sum_{n=1}^{\infty}\int_{r}^{1/r}\frac{e^{-nx}}{n+1}dx = \lim_{r\to+0} \sum_{n=1}^{\infty} [(\frac{1}{n+1}-\frac{1}{n})e^{-nx}]_{r}^{1/r}.$

No idea from here.

Answer 1

Note that $e^{-nx}\leq e^{-nr}=(e^{-r})^n.$ as $x\geq r$. Thus, any partial sum $S_n(x)$ will, uniformly in all $x\in [r,+\infty),$ satisfy $$S_n(x)\leq \sum_{k=1}^n \frac{(e^{-r})^k}{k+1}<\sum_{k=1}^n (e^{-r})^k.$$ The RHS is the partial sum of a Geometric series of common ratio $0<e^{-r}<1$--hence converges. Thus, your series converges uniformly.

2. because of uniform convergence, you can interchange the integral with the infinite sum. So we do this and after integration the result follows.