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Determine the convergence of the series with an = 1/n2 or an = -1 / n

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} a_n \]

where

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

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series diverges.

Proof. First, we write

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

We know the series

    \[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} \]

converges by comparison with \sum_{n=1}^{\infty} \frac{1}{n^2}. The series

    \[ \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \]

diverges by comparison with \sum_{n=1}^{\infty} \frac{1}{n}. Hence, the difference must diverge (since the sum/difference of a convergent series with a divergent series is divergent). \qquad \blacksquare

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