TITLE: The series $2+3x+5x^2+7x^3+11x^4+...$
QUESTION [7 upvotes]: It occurred to me to ask whether the power series whose coefficients are the primes has non-zero radius of convergence, and if so, what kind of function it produces.
Wikipedia has some bounds on the $n$th prime number, and assuming that they're correct, we eventually have (where the $^2$ indicates iteration):
$$p_n<n\log n + n\log^2n=n\log (n\log n)$$
Now, we have eventually $n < n\log n < n^2$, the left inequality as $\log n > 1$ and the right because $\log n < n$, and both logs and $n$th roots are increasing, so we conclude:
$$\sqrt[n]{\log n} < \sqrt[n]{\log(n\log n)} < \sqrt[n]{\log(n^2)}=\sqrt[n]2\sqrt[n]{\log n}$$
And as $1<\log n<n^2$, we have $\sqrt[n]{\log n}\to 1$ and so $\sqrt[n]{\log(n\log n)}\to 1$. So the series with $n$th coefficient $n\log (n\log n)$ has radius of convergence $1$. Since it's coefficient-wise greater than the series with prime coefficients, the latter has radius of convergence at least $1$.
Is this reasoning correct? Did I interpret the upper bound on $p_n$ correctly? (ie. is it exact, like I think it is, or is it just asymptotically true in some sense?) If I'm right, then what's known about the function $f(x)=\sum p_n x^n, x\in \mathbb C$? Otherwise, where did I go wrong, and is the radius of convergence actually non-zero?

REPLY [2 votes]: Indeed
$$
n< p_n<n \log (n\log n),
$$
implies that
$$
\lim_{n\to\infty}p_n^{1/n}=1,
$$
which by the root test implies that the radius of convergence of the power series is equal to $1$.