Show L'Hospital limit for exponential function and power series 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?
 A: Lets review first what you need to calculate your limit.
Obviously the power series
$$
f(t):=\sum_{k=0}^{\infty} \frac{t^{2k}}{\sqrt{(k!)}}
$$
Converges
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.
