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Determine the radius of convergence of ∑ (sinh (an)) zn

Determine the radius of convergence r of the power series:

    \[ \sum_{n=0}^{\infty} (\sinh (an)) z^n \qquad a > 0. \]

Test for convergence at the boundary points if r is finite.


First, we recall the definition of the hyperbolic sine in terms of the exponential,

    \[ \sinh x = \frac{e^x - e^{-x}}{2}. \]

Then we have,

    \begin{align*}  \sum_{n=0}^{\infty} ( \sinh (an))z^n &= \sum_{n=0}^{\infty} \left( \frac{e^{an} - e^{-an}}{2} \right) z^n \\[9pt]  &= \sum_{n=0}^{\infty} \left( \frac{e^{an} z^n}{2} - \frac{z^n}{2e^{an}} \right). \end{align*}

We know this series converges if and only if the two series

    \[ \sum_{n=0}^{\infty} \frac{e^{an}z^n}{2}, \qquad \text{and} \qquad \sum_{n=0}^{\infty} \frac{z^n}{2e^{an}} \]

converge. We know the second of these converges for |z| < \frac{2}{2e^a}. For the first we apply the root test,

    \[ a_n = \frac{e^{an} z^n}{2} \quad \implies \quad a_n^{\frac{1}{n}} = \frac{e^a z}{2^{\frac{1}{n}}}. \]

Therefore,

    \[ \lim_{n \to \infty} a_n^{\frac{1}{n}} = \lim_{n \to \infty} \frac{e^a z}{2^{\frac{1}{n}}} = e^a z. \]

Hence, we must have |z| < \frac{1}{e^a} which implies r = e^{-a}.

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