Suppose we are given a function $f$ in terms of its power series \begin{align} f(x)= \sum_{n+0}^\infty a_n x^n. \end{align} We assume that we know all $a_n$'s and $f$ has infinite radius of convergense.
Can we say based on $a_n$ if the function is integrable or not? That is if \begin{align} \int_{\mathbb{R}} |f(x)| dx<\infty. \end{align}
For example, if $a_n \ge 0$ then we have that \begin{align} f(x) \ge a_0+a_1 x,\ x >0 , \end{align} and the function is not integrable.
The case that I am interested is when $a_n$'s have alternating sign. Specifically, can we determine if the following $f(x)$ is integrable \begin{align} f(x)= \sum_{k=0}^\infty \frac{\cos( \frac{\pi}{2}k)\ (k+1)^{\frac{k+1}{6}}}{k!} x^k. \end{align}
Here is the plot of $f(x)$ where the series was computed up to $N=600$. 
where blue curve is $f(x)$ and red curve is $sinc(\pi/2 x)= \frac{\sin(\pi/2 x)}{\pi/2 x}$.
It seems that $f(x)$ has a tail that decays faster than $sinc(\pi/2 x)$.
Moreover, next we give a plot of $f(x) \cdot x$ and $f(x) \cdot x^2$
where blue curve is $f(x) \cdot x$ and red curve is $f(x) \cdot x^2$.
This seems to indicate that $f(x)$ decreases faster than $\frac{1}{x^2}$ which would imply that $f(x)$ is integrable.
Therefore, it would also be interesting to show that \begin{align} \lim_{x \to \infty} f(x)=0, \end{align} which at this moment I do not know how to do.
Thanks you.
**Edit: ** Please see a possible solution via Mellin transform.
To start consider mellin's inversion theorem.
Mellin's inversion theorem can be stated as, if
$$f(x) = \frac{1}{2\pi i}\int_{\sigma - i\infty}^{\sigma + i\infty}F(z)x^{-z}\,dz$$
Assuming everything is convergent, $0 < \sigma < 1$, $F$ holomorphic, then
$$F(z) = \int_0^\infty f(x)x^{z-1}\,dx$$
for $0 < \Re(z) < 1$
Ramanujan's master theorem gives us a really beautiful result based off this. If $z\Gamma(-z)g(z) \to 0$ as $z \to \infty$ when $|\arg(z)| < \pi/2$ and $g(z) = O(e^{\tau|\Im(z)|})$ for $\tau < \pi/2$, then
Ramanujan's Master theorem says
$$f(x) = \sum_{n=0}^\infty g(n)\frac{(-x)^n}{n!} = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma+ i \infty} \Gamma(z)g(-z)x^{-z}\,dz$$
If you need more justification of this fact look at Ramanujan's master theorem, even wikipedia has something on this. There are a few ways of stating this, but the version I'm using is perfectly valid.
What is great is that
$$\int_0^\infty |f(x)| x^{-\sigma}\,dx < \infty$$
Which is another consequence of Mellin's inversion theorem.
Now your function $F(z) = \cos(\pi/2 z)(z+1)^{(z+1)/6}$ satisfies these bounds. It is rather trivial to show this, if need be I'll edit this in. By the above
$$f(x) = \sum_{n=0}^\infty \cos(\pi/2n)(n+1)^{(n+1)/6}\frac{(-x)^n}{n!}$$
satisfies for $0 < \Re(z) < 1$
$$\int_0^\infty f(x)x^{z-1}\,dx = \Gamma(z)F(z)$$
Therefore off the start, your function $f$ is in a weighted $L^{1}$ space (with weight $x^{-\sigma}$ (this is also part of Mellin's inversion theorem). All the odd powers disappear, meaning the function is symmetric about zero. Therefore
$$\int_{-\infty}^\infty |f(x)|x^{-\sigma}\,dx < \infty$$
for all $ 0 < \sigma < 1$. I'll leave it to you to show the subtleties required in making it $L^1$. If you can't, I'll show it .