5
$\begingroup$

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!!

$\endgroup$
2
$\begingroup$

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

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

Thus

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

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.)

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $x\to0^+$, $\left(\frac1x-1\right)^ke^{-1}\to\infty$

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
-1
$\begingroup$

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

$\endgroup$
  • 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.