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?
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 $m/n$.
ADDENDUM
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.