# Limits, Taylor expansion

Find the limit: $$\lim_{x\to\infty} \frac\sum\limits_{i = 0}^\infty \frac{x^{in}}{(in)!}}{\displaystyle\sum\limits_{j = 0}^\infty \frac{x^{jm}}{(jm)!}}$$ 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?

• By $in!$ you mean $(in)!$, right? – enzotib May 27 '15 at 7:39
• I believe that this link might prove itself helpful. – Lucian May 27 '15 at 8:55
• Yeah, but I still can't find a general form for $\sum\limits_{i = 0}^\infty \frac{x^{in}}{(in)!}$ – prometheus21 May 27 '15 at 9:03

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.

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