# Radius of convergence of $\sum\limits_{n \ge 1} a_n z^n$ where $a_n$ is the number of divisors of $n^{50}$

Consider the power series $\sum_{n \ge 1} a_n z^n$, where $a_n$ is the number of divisors of $n^{50}$. What is its radius of convergence?

My attempt

$a_n < n^{50}$ $\forall$ $n$.

So $\lim \{a_n\}^{\frac{1}{n}} \le \{\lim n^{\frac{1}{n}}\}^{50} = 1$.

So radius of convergence is 1.

• en.wikipedia.org/wiki/Divisor_function#Approximate_growth_rate – Jean-Claude Arbaut Dec 11 '13 at 13:56
• Note that your attempt only shows that $\limsup a_n^{1/n} \le 1$, that is $R \ge 1$. To prove that $R=1$ you have to prove that also $\limsup a_n^{1/n} \ge 1$, that is find a sequence of $n$s with $a_n^{1/n} \to 1$ – martini Dec 11 '13 at 14:11
• @Marrini I know that you have pointed out, but I am not getting anything else to do. – Dutta Dec 11 '13 at 14:24
• @arbautjc Thank you for this link. I do not know number theory well. Is there any analytic approach to solve it? – Dutta Dec 11 '13 at 14:26
• My knowledge of number theory is very limited. – Dutta Dec 11 '13 at 15:09

Hint: For every $n$, the number of divisors of $n^{50}$ is between $1$ and $n^{50}$. Determine the radiuses of convergence of the series $\sum\limits_nz^n$ and $\sum\limits_nn^{50}z^n$. Conclude.

One sees that the proof uses neither the number of divisors of $n^{50}$, nor its exact growth, nor any other number theoretical refinement.

• Not an argument at all. Just saying that sometimes annoying somebody else helps relieve one's own annoyance. – Daniel Fischer Dec 11 '13 at 14:44
• This answer is more natural. I considered the series $\sum_n n^{50} z^n$. So there was only one series more to consider. So trivial !!! Thank you Did. – Dutta Dec 11 '13 at 15:00
• @DanielFischer THIS, I can fully understand (though, not endorse...). :-) – Did Dec 11 '13 at 15:04

You are only proving that the radius of convergence is $≥1$.

When you observe that for $n=2^k$ the number of divisors of $n$ is $2^{k-1}$, so the number of divisors of $n^{50} = 2^{50k}$ is $2^{50k-1}$, then you can work out the missing inequality.

Edit: the answer is wrong. $d(2^k) = k+1$ so it doesn't prove anything. Actually the growth rate of $d(n)$ is very low, lower than any positive power of $n$ (still greater than $\log n$)

• The number of divisors of $2^k$ (more generally $p^k$ for any prime $p$) is $k+1$, not $2^{k-1}$. – Daniel Fischer Dec 11 '13 at 14:31
• @DanielFischer whops.... – rewritten Dec 11 '13 at 14:57