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Determine the radius of convergence of ∑ (an / n + bn / n2)zn

Determine the radius of convergence r of the power series:

    \[ \sum_{n=1}^{\infty} \left( \frac{a^n}{n} + \frac{b^n}{n^2} \right) z^n, \]

where a> 0 and b > 0. Test for convergence at the boundary points if r is finite.


First, we write

    \[ \sum_{n=1}^{\infty} \left( \frac{a^n}{n} + \frac{b^n}{n^2} \right) z^n = \sum_{n=1}^{\infty} \frac{(az)^n}{n} + \sum_{n=1}^{\infty} \frac{(bz)^n}{n^2} \]

where the sum on the left converges if and only if both sums on the right converge (since a> 0 and b> 0 these are all positive terms; hence, we cannot have their sum convergent while either of the individual series are divergent since they would diverge to +\infty). So, we apply the root test to the two series on the right:

    \begin{align*}  \lim_{n \to \infty} \left( \frac{(az)^n}{n} \right)^{\frac{1}{n}} &= \lim_{n \to \infty} \frac{az}{n^{\frac{1}{n}}} = az \\[9pt]  \lim_{n \to \infty} \left( \frac{(bz)^n}{n^2} \right)^{\frac{1}{n}} &= \lim_{n \to \infty} \frac{bz}{n^{\frac{2}{n}}} = bz. \end{align*}

Thus, the first series on the right converges for |z| < \frac{1}{a} and the second converges for |z| < \frac{1}{b}. Therefore, the series on the left converge for

    \[ |z| < \frac{1}{\max(a,b)} \quad \implies \quad r = \frac{1}{\max (a,b)} \quad \implies \quad r = \min \left( \frac{1}{a}, \frac{1}{b} \right). \]

2 comments

  1. tom says:

    I think the boundary points converge as well- either all points by comparison with 1/n^2, or all z≠1 by dirichlet depending on b or a being larger, respectively.

    • Matt says:

      The boundary points do converge. The following outline explains Tom’s comment a bit more:\\
      For the case where b > a, let |z| = \frac{1}{b} and compare the magnitude of the series with \Sigma \frac{1}{n^2}. The series absolutely converges on the boundary.\\
      For the case where b \le a, let |z| = \frac{1}{a} and use Dirichlet’s test along with theorem 10.19, to show that the series converges for z \ne 1.\\

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