$\{p_n\}$ is a convergent sequence while $\sum_{n=1}^{\infty}|p_n − p_{n+1}|$ diverges.

Question

Give an example of a convergent sequence $\{p_n\}$ in $\mathbb R$ such that $\sum_{n=1}^{\infty}|p_n − p_{n+1}|$ diverges.

I had another question that was the 'opposite' of this i.e. $p_n$ diverges and the difference converges which I had little trouble completing using Cauchy, but I'm drawing a blank on this example.

Answer 1

For example $p_n=(-1)^n\frac{1}{n}$ converges by the alternating series test. $$\sum_{n=1}^{\infty} |p_n-p_{n+1}|=\sum_{n=1}^{\infty} \left(\frac{1}{n}+\frac{1}{n+1}\right)>2\sum_{n=1}^{\infty} \frac{1}{n}$$which diverges

Answer 2

For your first question, something like

$$\left\{0, \frac 12 , 0, \frac 13, 0, \frac 14, 0, \cdots\right\}$$

would do. For the opposite question, if $$\tag{1} \sum_{n=1}^\infty |p_n - p_{n+1}|$$ converges, then $\{p_n\}$ is a Cauchy sequence (thus is convergent). To see this, let $\epsilon >0$. Then as $(1)$ converges, there is $N_\epsilon \in \mathbb N$ so that $$\sum_{n = N_\epsilon} ^\infty |p_n - p_{n+1}| <\epsilon.$$ Then for all $n, m\ge N_\epsilon$, $n<m$, we have $$\begin{split}|p_n - p_m| &= |p_n - p_{n+1} + p_{n+1} - p_{n+2} + \cdots -p_{m-1} + p_{m-1} -p_m|\\ &\le |p_n - p_{n+1}| + |p_{n+1} - p_{n+2}|+ \cdots + |p_{m-1}- p_m|\\ & \le \sum_{k=n}^{m-1} |p_k- p_{k+1}| <\epsilon. \end{split}$$ Thus $\{p_n\}$ is Cauchy.