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

by RoRi

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}.

One comment

  1. Artem says:

    There is a mistake in the statement – when the sums of series concerned, it is not “if and only if”. The example is series of -1 and 1. The sum is -1 + 1 = 0, it converges.Separately they diverge. Thus, the resultant sum can converge either if both summands converge, or if both summands diverge. The second case can be checked by assuming the radius of convergence >1 for both, then checking the case r = 1, finding counterexample, and then invoking theorem 11.6, which shows that both summands have to converge.

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