4
$\begingroup$

Let $[a_1,a_2,a_3,...,a_n]$ be the least common multiple of numbers $a_1,a_2,...,a_n$. Then what should the radius of convergence be for the following series:$$\sum_{n=1}^{\infty} \frac{z^n}{[1,2,...,n]}$$

$\endgroup$
5
$\begingroup$

By the prime number theorem, the Chebyshev function has the asymptotic $\psi(n) = n(1+o(1))$ , which tells that $[1, 2, \ldots, n] = \exp(\psi(n)) = \exp(n(1+o(1))$. Therefore, $[1, 2, \ldots, n]^{1/n} = \mathrm e+o(1)$, and hence the radius of convergence is $\mathrm e$.

$\endgroup$
2
$\begingroup$

Hint: OEIS says $[1,2,...,n]$ goes asymptotically as $\exp(n)$

$\endgroup$

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.