Let $\sum_{n=1}^{\infty} a_n z^{n}$ be a power series with $a_n$ equal to number of divisors of $n^{50}$. How can I find the radius of convergence.
My efforts.
$\beta=\limsup_{n \rightarrow \infty}|a_n/a_{n+1}|$. Using this I got the answer 1.
But the books says answer is zero. I am confused. I think there is a printing mistake.
Since each $a_n$ is at least $1$, the series diverges when $z=1$ and therefore the radius of convergence is at most $1$.
On the other hand, $n^{50}$ has at most $n^{50}$ divisors and therefore $a_n\leqslant n^{50}$. So,$$\limsup\sqrt[n]{a_n}\leqslant\limsup_n\sqrt[n]{n^{50}}=1.$$Therefore, the radius of convergence is at least $1$.
Conclusion: …