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.