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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:$$\sum_{n=1}^{\infty} \frac{z^n}{[1,2,...,n]}$$

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up vote 5 down vote accepted

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 $\mathrm e$.

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

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