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Test convergence of the series with an = 1/n if n odd and an = 1/n2 if n even

by RoRi

Test the series \sum_{n=1}^{\infty} a_n for convergence where the terms a_n are defined by

    \[ a_n = \begin{cases} \frac{1}{n} & \text{if } n \text{ is odd}, \\ \frac{1}{n^2} & \text{if } n \text{ is even}. \end{cases} \]

The series diverges.

Proof. (This proof might be overkill. You could simply observe that this series is greater than the sum of the reciprocals of the odd integers, and use limit comparison with \frac{1}{n}.)Using the piecewise definition of a_n we have

    \[ \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} + \frac{1}{(2n)^2} \right) = \sum_{n=1}^{\infty} \frac{4n^2 + 2n - 1}{8n^3 - 4n^2}. \]

Then using the limit comparison we have,

    \begin{align*} \lim_{n \to \infty} \frac{ \frac{4n^2 + 2n - 1}{8n^3 - 4n^2}}{\frac{1}{n}} &= \lim_{n \to \infty} \frac{4n^3 + 2n^2 - n}{8n^3 - 4n^2} \\[9pt] &= \frac{1}{2}. \end{align*}

Since we know \sum \frac{1}{n} diverges, the limit comparison test establishes the divergence of

    \[ \sum a_n. \qquad \blacksquare \]

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