counterexamples-around-series-part-1-image

Counterexamples around series (part 1)

The purpose of this article is to provide some basic counterexamples on real series. Counterexamples are provided as answers to questions.

Unless otherwise stated, \((u_n)_{n \in \mathbb{N}}\) and \((v_n)_{n \in \mathbb{N}}\) are two real sequences.

If \((u_n)\) is non-increasing and converges to zero then \(\sum u_n\) converges?

Is not true. A famous counterexample is the harmonic series \(\sum \frac{1}{n}\) which doesn’t converge as \[
\displaystyle \sum_{k=p+1}^{2p} \frac{1}{k} \ge \sum_{k=p+1}^{2p} \frac{1}{2p} = 1/2,\] for all \(p \in \mathbb N\).

If \(u_n = o(1/n)\) then \(\sum u_n\) converges?

Does not hold as can be seen considering \(u_n=\frac{1}{n \ln n}\) for \(n \ge 2\). Indeed \(\int_2^x \frac{dt}{t \ln t} = \ln(\ln x) – \ln (\ln 2)\) and therefore \(\int_2^\infty \frac{dt}{t \ln t}\) diverges. We conclude that \(\sum \frac{1}{n \ln n}\) diverges using the integral test. However \(n u_n = \frac{1}{\ln n}\) converges to zero.

If \(u_n = o(v_n)\) and \(\sum v_n\) converges then \(\sum u_n\) converges?

Is not true as we can see taking \(v_n = \frac{(-1)^n}{n}\) and \(u_n=\frac{1}{n \ln n}\). \(\sum v_n\) converges according to the alternating series test and \(\sum u_n\) diverges according to previous paragraph. However \[
\lim\limits_{n \to \infty} \left\vert \frac{u_n}{v_n} \right\vert = \lim\limits_{n \to \infty} \frac{1}{\ln n} = 0.\]

If \(\sum u_n\) is a positive converging series then \(\frac{u_{n+1}}{u_n}\) is bounded?

Not either. Have a look at \[
u_n=\begin{cases} \frac{1}{n^2} & \text{for } n \text{ even} \\
\frac{1}{2^n} & \text{for } n \text{ odd.} \end{cases}\] Then for \(p \in \mathbb N\) \[
\frac{u_{2p+1}}{u_{2p}} = \frac{2^{2p+1}}{4p^2}\] diverges to \(\infty\). However \(\sum u_n\) converges as \(\sum \frac{1}{n^2}\) and \(\sum \frac{1}{2^n}\) both converge.

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