Does this function grow slower than $e^{x}$ but faster than $e^{kx}$ for all $0<k<1$

Question

$f\left(x\right)=\lim_{a\to\infty}\sum_{n=1}^{a}\frac{x^{n}}{n\cdot n!}$

I spontaneously thought of this function while messing with the power series for exp(x). It's fairly obvious that this function grows slower than exp(x) and evaluating further derivatives at x=0 seems to indicate that it grows faster than all $e^{kx}$ functions (for 0<k<1), as their derivatives follow $k$,$k^2$,$k^3$, etc, while derivatives of f(x) are 1, 1/2, 1/3 etc. The former derivatives shrink faster than the latter. I think this is not really a valid argument (since it's all at zero), but my guess seems to be true from my observations. Is it, and how would you prove it in that case?

edit: by "grows faster" I guess I mean as x→∞, the ratio of f(x) / $e^{kx}$ approaches infinity.

P.S. I apologise if this is a well-known function or if I screwed up the latex formatting.

Answer 1

We have $$f(x) =\sum_{n=1}^\infty \frac{x^n}{n\cdot n!} \sim e^x/x$$ so your conjecture is true.

To prove the asymptotic equivalence observe that $$f(x) = \int_0^x\frac{e^t-1}{t}\ dt$$ and we can calculate the following limit using L'Hôpital's rule:

$$\lim_{x\to\infty}\frac{f(x)}{e^x/x} = \lim_{x\to\infty}\frac{\frac{e^x-1}{x}}{\frac{e^x x-e^x}{x^2}}=\lim_{x\to\infty}\frac{e^x-1}{e^x}\cdot\frac{x}{x-1}=1$$

Answer 2

In general if $$f(x)=\sum a_nx^n,\quad g(x)=\sum b_nx^n$$ are convergent on the whole line, with $a_n,b_n>0$ for $n\ge n_0$ and $b_n/a_n\to 0,$ then $$\lim_{x\to\infty}{f(x)\over g(x)}=\infty\quad (*)$$ Indeed, for $\delta>0$ there is $n_\delta> n_0$ such that $0<b_n/a_n<\delta$ for $n\ge n_\delta.$ Then for $x>0$ we get $$g(x)=\sum_{n=0}^{n_\delta-1}b_nx^n+\sum_{n=n_\delta}b_nx^n\\ \le\sum_{n=0}^{n_\delta-1}b_nx^n +\delta\sum_{n=n_\delta}a_nx^n\\ =\sum_{n=0}^{n_\delta-1}b_nx^n +\delta\ f(x)$$ Thus $$0<{g(x)\over f(x)}\le f(x)^{-1}\sum_{n=0}^{n_\delta-1}b_nx^n+\delta\\ \le a_{n_\delta}^{-1}x^{-n_\delta}\sum_{n=0}^{n_\delta-1}b_nx^n+\delta $$ The first term tends to $0$ when $x\to \infty $ thus it is less than $\delta$ for $x>x_\delta$ for some $x_\delta>0.$ Hence $${g(x)\over f(x)}\le 2\delta,\quad x>x_\delta $$ This completes the proof of $$\lim_{x\to \infty}{g(x)\over f(x)}=0$$ The latter implies $(*).$

In OP case we have $$a_n={1\over n\,n!},\quad b_n={k^n\over n!},\ 0<k<1$$ Hence $a_n/b_n={1/(nk^n)}\to \infty.$