In that article, I provide basic counterexamples on sequences convergence. I follow on here with some additional and more advanced examples.

If \((u_n)\) converges then \((\vert u_n \vert )\) converge?

This is true and the proof is based on the reverse triangle inequality: \(\bigl| \vert x \vert – \vert y \vert \bigr| \le \vert x – y \vert\). However the converse doesn’t hold. For example, the sequence \(u_n=(-1)^n\) is such that \(\lim \vert u_n \vert = 1\) while \((u_n)\) diverges.

If for all \(p \in \mathbb{N}\) \(\lim\limits_{n \to +\infty} (u_{n+p} – u_n)=0\) then \((u_n)\) converges?

The assertion is wrong. A simple counterexample is \(u_n= \ln(n+1)\). It is well known that \((u_n)\) diverges. However for any \(p \in \mathbb{N}\) we have \(\lim\limits_{n \to +\infty} (u_{n+p} – u_n) =\ln(1+\frac{p}{n+1})=0\).
The converse proposition is true. Assume that \((u_n)\) is a converging sequence with limit \(l\) and \(p \ge 0\) is any integer. We have \(\vert u_{n+p}-u_n \vert = \vert (u_{n+p}-l)-(u_n-l) \vert \le \vert u_{n+p}-l \vert – \vert u_n-l \vert\) and both terms of the right hand side of the inequality are converging to zero.

If \(\lim (u_{2n} – u_n) = 0\) and \((u_n)\) is increasing then \((u_n)\) converges?

Is not true. However, in order to find a counterexample, we have to pick-up a slowly increasing sequence. As a matter of fact the sequence \(\ln(n+1)\) is increasing too quick as \(\lim \bigl(\ln(2n+1) – \ln(n+1)\bigr) = \ln 2\). The sequence \(u_n=\ln(\ln(n+1))\) provides a counterexample. \(u_n\) is increasing as the function composition of two increasing maps. We also have the equality:
\[u_{2n} – u_n=\ln \left(\frac{\ln 2}{\ln(n+1)}+\frac{\ln(n+1/2)}{\ln(n+1)}\right)\] and the right end side converges to \(0\).

If \((u_n)\) is positive and unbounded then \(u_n \to +\infty\)?

Does not hold. Have a look at \(u_n=\begin{cases} 0 & \text{for } n \text{ even} \\
n & \text{for } n \text{ odd} \end{cases}\)

If \(\lim u_n = +\infty\) then \((u_n)\) is eventually increasing?

Still not! The sequence \(u_n=\begin{cases} n & \text{for } n \text{ even} \\
2n & \text{for } n \text{ odd} \end{cases}\) provides a counterexample.

If \(\lim (u_{n+1} – u_n) = 0\) then \(\lim \frac{u_{n+1}}{u_n} = 1\)?

Is also wrong as you can see with the sequence \(u_n=\begin{cases} \frac{1}{n+1} & \text{for } n \text{ even} \\
\frac{1}{(n+1)^2} & \text{for } n \text{ odd.} \end{cases}\)

If \(\lim (u_{n+1}-u_n) =0 \) and \((u_n)\) is bounded then \((u_n)\) converges?

This is still wrong! But it is slightly more difficult to find a counterexample. One idea is to find a sequence which oscillates between \(0\) and \(1\) with steps decreasing to zero. The sequence:
\[u_n=\begin{cases} 0 & \text{for } n = 0\\
1-\frac{n-2^{2k}}{2^{2k}} & \text{for } 2^{2k} \le n < 2^{2k+1}\\ \frac{n-2^{2k+1}}{2^{2k+1}} & \text{for } 2^{2k+1} \le n < 2^{2k+2}\end{cases}\] is well defined and having all the required properties:

• for all \(n \in \mathbb{N}\), \(u_n \in [0,1]\), therefore \((u_n)\) is bounded.
• For \(n \ge 2^k\) we have \(\vert u_{n+1}-u_n \vert \le \frac{1}{2^k}\), hence \(\lim (u_{n+1}-u_n) =0 \).
• However, \((u_n)\) diverges as for all \(k \in \mathbb{N}\) we have \(u_{2^{2k}}=1\) and \(u_{2^{2k+1}}=0\).

It is interesting to notice that the set of limit points of \((u_n)\) is the interval \([0,1]\).

The examples above were inspired from mathematical exercises given by math teacher Robert FERRÉOL.