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!

  • 3
    $\begingroup$ In step 2, you need a majorisation that works (for all [large enough] $k$) simultaneously for all $n$. Can you show that $\frac{k}{n}e^{-k/n} \leqslant \frac{1}{2}$ for all $k$ and $n$? [Essentially, you want an upper bound for the function $x\mapsto xe^{-x}$ on $[0,+\infty)$.] $\endgroup$ – Daniel Fischer Dec 31 '16 at 16:37
  • 1
    $\begingroup$ Note that $\frac kne^{-k/n}\le e^{-1}$. $\endgroup$ – Mark Viola Dec 31 '16 at 16:49
  • $\begingroup$ I thought that the majorisation was allowed to depend on n? $\endgroup$ – Staki42 Jan 1 '17 at 16:54
  • $\begingroup$ If the majorization were allowed to depend on $n$, then the fact that each series converges would provide a majorization so that we can switch the order of limit and sum for any sequence of absolutely convergent series, and that is not the case; for example, using Iverson brackets, $$\lim_{n\to\infty}\sum_{k=1}^\infty\frac{[k\le n]}n=1$$ but $$\sum_{k=1}^\infty\lim_{n\to\infty}\frac{[k\le n]}n=0$$ $\endgroup$ – robjohn Feb 11 '20 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.