6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ en.wikipedia.org/wiki/Divisor_function#Approximate_growth_rate $\endgroup$ – Jean-Claude Arbaut Dec 11 '13 at 13:56
  • $\begingroup$ 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 $a_n^{1/n} \to 1$ $\endgroup$ – martini Dec 11 '13 at 14:11
  • $\begingroup$ @Marrini I know that you have pointed out, but I am not getting anything else to do. $\endgroup$ – Dutta Dec 11 '13 at 14:24
  • $\begingroup$ @arbautjc Thank you for this link. I do not know number theory well. Is there any analytic approach to solve it? $\endgroup$ – Dutta Dec 11 '13 at 14:26
  • $\begingroup$ My knowledge of number theory is very limited. $\endgroup$ – Dutta Dec 11 '13 at 15:09
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Not an argument at all. Just saying that sometimes annoying somebody else helps relieve one's own annoyance. $\endgroup$ – Daniel Fischer Dec 11 '13 at 14:44
  • $\begingroup$ 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. $\endgroup$ – Dutta Dec 11 '13 at 15:00
  • $\begingroup$ @DanielFischer THIS, I can fully understand (though, not endorse...). :-) $\endgroup$ – Did Dec 11 '13 at 15:04
1
$\begingroup$

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 $\log n$)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.