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

by RoRi

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

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

3 comments

  1. Anonymous says:

    In case of b > a > 0, bound points with \lvert z \rvert = \frac {1} {b}

        \[\lvert \frac {a^n} {n} z^n\rvert = (\frac {a} {b}) ^n \\ \lvert \frac {b^n} {n^2} z^n\rvert = \frac {1} {n^2}\]

    so, \sum \lvert \frac {a^n} {n} z^n\rvert and \sum \lvert \frac {b^n} {n^2} z^n\rvert converge.
    And also,

        \[\lvert \frac {a^n} {n} z^n + \frac {b^n} {n^2} z^n \rvert 0$, boundary points with $\lvert z \rvert = \frac{1}{a}$\]

    \lvert \frac {a^n} {n} z^n + \frac {b^n} {n^2} z^n \rvert = \lvert \frac{1}{n} + \frac{(\frac{b}{a})^n}{n^2} \rvert = \frac{1}{n} + \frac{(\frac{b}{a})^n}{n^2} > \frac{1}{n}

        \[Because $\sum \frac{1}{n}$ diverge\]

    \sum \lvert \frac {a^n} {n} z^n + \frac {b^n} {n^2} z^n \rvert

    diverge

  2. 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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