Given a series $$f(t):=\sum_{k=0}^{\infty} \frac{t^{2k}}{\sqrt{(k!)}},$$

then since by first term expansion we have $f(t)\ge 1+t^2$, we get that $f(t) \rightarrow \infty$ for $t \rightarrow \infty.$

But now I want to compare the growing rate of this function to the expoentnial one and show that for any $a>0$ we have:

$$\frac{\text{exp}(a\ t^a)}{f(t)} \rightarrow 0$$ for $t \rightarrow \infty$. Is there a way to show this completely rigorously?

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    $\begingroup$ For $a<2$, this can be shown by $f(t) \ge \exp(t^2)$. $\endgroup$ – Hetebrij Apr 16 '16 at 9:40
  • $\begingroup$ @Hetebrij how exactly does $f(t) \ge exp(t^2)$ imply this? $\endgroup$ – user331288 Apr 16 '16 at 12:35
  • $\begingroup$ As $at^a -t^2 \to - \infty$ when $t \to \infty$ you have $\exp( at^a -t^2) \to 0$. $\endgroup$ – Hetebrij Apr 16 '16 at 14:46

Lets review first what you need to calculate your limit.

Obviously the power series

$$ f(t):=\sum_{k=0}^{\infty} \frac{t^{2k}}{\sqrt{(k!)}} $$


since lets look at

$$ g(x):=\sum_{k=0}^{\infty} \frac{x^{k}}{\sqrt{(k!)}} $$

Its obvious that $f(t)=g(t^2)$ And looking at $g(x)$ we see from $$ \lim_{k\rightarrow +\infty} \frac{a_k}{a_{k+1}}=+\infty $$

That $g(x)$ converges from $(-\infty,+\infty)$

That said we know $f(t)$ converges and we also know that we can differentiate term by term and that it such a series converges in the interior of $(-\infty,+\infty)$ (so it converges on $(-\infty,+\infty)$

Ok so we know that now we can say that

$$ f'(t)=\sum_{k=0}^{\infty} \frac{2kt^{2k-1}}{\sqrt{(k!)}} $$

However this leads us nowhere (at least not that i see)

We will use that $$ (e^{at^{a}})=\sum_{k=0}^{+\infty} \frac{a^k t^{ak}}{k!} $$

Note that that infinite sum also converges for $t\in (-\infty,+\infty)$ And that therefore we can add/subtract those two sums.

So we know that

$$ f(t)-exp(at^a)=\sum_{k=0}^{+\infty} \frac{\sqrt{k!}t^{2k}-a^kt^{ak}}{k!} $$

For big enough k we will have that $k>a $ hence $t^2k > t^{ak}$ and also $\sqrt{k!} > a^k$ again starting from some k.

From there i think you can solve the rest.

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