TITLE: Radius of convergence of $\sum a_nz^n$, $a_n$ the number of divisors of $n^{50}$
QUESTION [8 upvotes]: I have to find the radius of convergence of the power series $\sum a_n z^n$ where $ a_n =$ number of divisors of $n^{50}$. 
Options available are:

$1$
$50$
$\frac{1}{50}$
$0$

Please suggest how to proceed. 
Using the fact that $ d(n)\leq n$, we have $d(n^{50})\leq n^{50}$.
Using the First Comparison Test, the series on the right converges if $\mid{z}\mid< 1$ and however does not converge if $\mid{z}\mid\geq 1$ and hence so does the series on the left. 
Radius of Convergence is 1.  
[Now for $a_n =n$, then $ R = \lim_{n\rightarrow\infty}\frac {a_n}{a_{n+1}} =1. $
So,the series on the right converges for $\mid z \mid<1$]
Now if $ z=1$, then the series $\sum a_n z^n$ takes the form $\sum d(n^{50})$ and since
$ d(n^{50})\geq 1 $, by evoking the comparison test again, the series diverges for $z=1$.

REPLY [5 votes]: Hint: Use the standard inequality $d(n) \leq n$ to compare the coefficients of this series with another one.  Obtain a lower bound on the radius of convergence.
Hint 2: Once you have found a lower bound on the radius of convergence, test the convergence at a point on the boundary to show that the bound you found above actually is the exact radius of convergence.
(Here $d(n)$ is the number of divisors of $n$.)