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Consider the power series $\sum\limits_{n=1}^{\infty}$$a_nZ^n$ , where $a_n$ is the number of divisor of $n^{50}$ . Find the radius of convergence.

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this sounds interesting... where did you find this problem?? have you tried something with this? – Praphulla Koushik Nov 29 '13 at 4:58

Use the facts that

  1. $\tau(n) \le 2\sqrt{n}$ for all $n$
  2. For infinitely many $n$ (say powers of $2$), $\tau(n) \ge \log_{2} n$
  3. $n^{1/n} \to 1$ and $(\log n)^{1/n} \to 1$
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We know that number of divisor of $n^{50}$ is $\tau$($n^{50}$) which tends to $\infty$ as n tends to $\infty$ . Let R be the radius of convergence, then $\frac{1}{R}$ = $\lim_{n\to \infty}$ sup $(a_n)^{\frac{1}{n}}$ is $\infty$, so R is 0. So please tell me am I right or not?

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Don't think it is true. – Aryabhata Nov 29 '13 at 9:07
(And please make this part of the question, if you are interested in the validity) – Aryabhata Nov 29 '13 at 9:15

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