I am trying to evaluate the following limit: $$ \lim_{x \to 0} \frac{e^x}{\sum_{n = 1}^\infty n^k e^{-nx}}, $$ where $k$ is a large (but fixed) positive integer. I am unsure how to proceed. Can this be done using L'Hopital's rule? Just started learning calculus, thanks guys!!


For any $N\in \mathbb N,$ we have for $x>0$ that

$$\frac{e^x}{\sum_{n=1}^{\infty}n^ke^{-nx}} < \frac{e^x}{N^ke^{-Nx}}.$$


$$0\le \limsup_{x\to 0^+} \frac{e^x}{\sum_{n=1}^{\infty}n^ke^{-nx}} \le \limsup_{x\to 0^+}\frac{e^x}{N^ke^{-nx}} = \lim_{x\to 0} \frac{e^x}{N^ke^{-Nx}} = \frac{1}{N^k}.$$

Since $N$ is arbitrary, the $\limsup$ on the left is arbitrarily small, hence equals $0.$ Thus the limit is $0$ from the right.

(Note that to the left of $0$ the sum in the denominator of our expression is identically $\infty.$ Not sure what to make of that.)

  • $\begingroup$ Shouldn't $n$ in the right-hand side of the first inequality be $N$? $\endgroup$ – egreg Feb 4 '16 at 22:47
  • $\begingroup$ Yes it should, thanks. $\endgroup$ – zhw. Feb 5 '16 at 2:26

For each $x\lt\frac12$, there is an $n\in\mathbb{N}$ so that $1\lt\frac1x-1\le n\lt\frac1x$. Picking this one term out gives $$ \sum_{n=0}^\infty n^ke^{-nx}\ge \left(\frac1x-1\right)^ke^{-1} $$ As $x\to0^+$, $\left(\frac1x-1\right)^ke^{-1}\to\infty$


I think, the whole converges to 0 (zero), because the numerator converges to 1 and the denominator converges to infinity.

  • $\begingroup$ Limit of a series = series of the limits? Not always. $\endgroup$ – user228113 Feb 4 '16 at 22:14
  • $\begingroup$ @G. Sassatelli : invertion of limits is always allowed when the sequence is increasing in the two directions. $\endgroup$ – reuns Feb 4 '16 at 22:23
  • $\begingroup$ @user1952009 My bad, I thought this used the same argument as the other one, while this actually cites "convergence". Fine to me, as long as other people understand. $\endgroup$ – user228113 Feb 4 '16 at 22:31

You don't need L'Hopital's rule; try substituting $0$ for $x$ and see what you get.

  • 1
    $\begingroup$ Limit of a series = series of the limits ? Not always, though I admit that using l'Hopital as suggested in the OP would at least have the same issue. $\endgroup$ – user228113 Feb 4 '16 at 22:17
  • $\begingroup$ The series at the denominator doesn't converge for $x=0$, but it converges for $x>0$, so plugging in $x=0$ is not allowed. (Actually the limit in the question only makes sense for $x\to0^+$.) $\endgroup$ – egreg Feb 4 '16 at 22:46

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