The following exercise is taken from a Calculus I course exam:
Let $k\in \mathbb N$. Prove the existence of $$x = \lim_{k\to \infty}\sum_{n=1}^{\infty}\exp\left(-n+\frac{k}{n}e^{-\frac{k}{n}}\right)$$ and calculate $x$.
This is an obvious application of the Dominated convergence theorem for infinite series. I think I found a solution, however I am not sure if it is entirely correct:
Step 1: We have to show that for all $n\in \mathbb N$ the limit $\lim_{k\to \infty}\exp\left(-n+\frac{k}{n}e^{-\frac{k}{n}}\right)$ exists. This holds since, for all $n\in \mathbb N$: $$\lim_{k\to \infty}\exp\left(-n+\frac{k}{n}e^{-\frac{k}{n}}\right) = \lim_{k\to \infty}\exp(-n) \cdot \lim_{k \to \infty}\exp\left(\frac{k}{n}e^{-\frac{k}{n}}\right) = \exp(-n),$$ having used that the exponential function grows faster than every polynomial.
Step2: We have to show the existence of a majorant sequence. Since $\frac{k}{n}e^{-\frac{k}{n}}$ converges towards $0$ for all $n\in \mathbb N$ we can choose $k\in \mathbb N$ so big that $\frac{k}{n}e^{-\frac{k}{n}} \leq \frac{1}{2}$ (or any other finite number). It thereby follows that $$\exp\left(-n+\frac{k}{n}e^{-\frac{k}{n}}\right) = \exp(-n) \cdot \exp\left(\frac{k}{n}e^{-\frac{k}{n}}\right) \leq \exp(-n) \cdot \exp\left(\frac{1}{2}\right) =: b_n$$ and $\sum b_n$ obviously converges (geometric series).
Now we can apply the Dominated convergence theorem yielding the existence of $x$. Furthermore, we get $$x = \lim_{k\to \infty}\sum_{n=1}^{\infty}\exp\left(-n+\frac{k}{n}e^{-\frac{k}{n}}\right) = \sum_{n=1}^{\infty}\lim_{k\to \infty}\exp\left(-n+\frac{k}{n}e^{-\frac{k}{n}}\right) = \sum_{n=1}^{\infty}\exp(-n) \quad\quad\quad= \frac{1}{e}\sum_{n=0}^{\infty}\exp(-n) = \frac{1}{e-1}$$
Is my approach correct or does it have flaws? Step 2 is the one I am not happy with but don't know how to do it better. Thanks in advance for your feedback!