Find the limit: $$ \lim_{x\to\infty} \frac{\displaystyle \sum\limits_{i = 0}^\infty \frac{x^{in}}{(in)!}}{\displaystyle\sum\limits_{j = 0}^\infty \frac{x^{jm}}{(jm)!}} $$ for $n$, $m$ natural numbers.

It is easy to see that for elementary cases, like $n=0$ we just get the Taylor expansion for $e^x$. We get the limit equal to $1$ for $n=m$. Any idea how to find a general rule? Maybe we can use some sort of squeezing argument?

  • $\begingroup$ By $in!$ you mean $(in)!$, right? $\endgroup$
    – enzotib
    May 27, 2015 at 7:39
  • $\begingroup$ I believe that this link might prove itself helpful. $\endgroup$
    – Lucian
    May 27, 2015 at 8:55
  • $\begingroup$ Yeah, but I still can't find a general form for $\sum\limits_{i = 0}^\infty \frac{x^{in}}{(in)!}$ $\endgroup$ May 27, 2015 at 9:03

1 Answer 1


The answer is $m/n$. The reason is that

$$f(n,x) = \sum_{j=0}^{\infty} \frac{x^{j n}}{(j n)!} = \frac1{n} \sum_{k=0}^{n-1} \exp{\left ( e^{i 2 \pi k/n} x\right )} $$

The sum is dominated by the $k=0$ term as $x \to \infty$. The ratio of such terms is thus $m/n$.


Proof of the above assertion is straightforward. The Taylor expansion of the RHS is

$$\frac1{n} \sum_{k=0}^{n-1} \sum_{j=0}^{\infty} \frac{e^{i 2 \pi j k/n} x^j}{j!} $$

Reverse order of summation (justified because each individual sum absolutely converges):

$$\frac1{n} \sum_{j=0}^{\infty} \sum_{k=0}^{n-1} \frac{e^{i 2 \pi j k/n} x^j}{j!} = \frac1{n} \sum_{j=0}^{\infty}\frac{ x^j}{j!} \sum_{k=0}^{n-1} e^{i 2 \pi j k/n}$$

The inner sum is a geometrical series, so the Taylor expansion is now

$$ \frac1{n} \sum_{j=0}^{\infty}\frac{ x^j}{j!} \frac{e^{i 2 \pi j} - 1}{e^{i 2 \pi j/n} - 1} $$

It should be clear that the latter factor is equal to zero unless $j$ is equal to a multiple of $n$, where it is equal to $n$. QED.

  • $\begingroup$ you put exp, so I think you don't need the $e$ $\endgroup$ May 27, 2015 at 10:08
  • 1
    $\begingroup$ @prometheus21: the two exponentials are required. $\endgroup$
    – user65203
    May 27, 2015 at 10:14
  • $\begingroup$ Yes I see that now $\endgroup$ May 27, 2015 at 10:15
  • $\begingroup$ Very clever way to keep only the the $in^{th}$ terms ! $\endgroup$
    – user65203
    May 27, 2015 at 10:15
  • $\begingroup$ @YvesDaoust: thanks. $\endgroup$
    – Ron Gordon
    May 27, 2015 at 10:24

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