Does $|a_{k+1}|\leq|a_k|$ and $\sum_{k=0}^\infty|a_{k+1}-a_{k}|^2<\infty$ imply $\sum_{k=0}^\infty|a_{k+1}-a_{k}|<\infty$?

Question

For all $k\in\mathbb{N}$, let $a_k\in\mathbb{R}$, and assume that $|a_{k+1}|\leq|a_k|$. Moreover, assume that $\sum_{k=0}^\infty|a_{k+1}-a_{k}|^2$ exists.

Can we prove or disprove by counterexample that $\sum_{k=0}^\infty|a_{k+1}-a_{k}|$ exists?

Answer 1

No.

For example $a_n=\dfrac{(-1)^n}{n}$. Then $$ |a_n-a_{n+1}|=\frac{1}{n}+\frac{1}{n+1}=\frac{2n+1}{n(n+1)}, $$ and hence $$ \frac{2}{n+1}<|a_n-a_{n+1}|<\frac{2}{n}\quad\text{and}\quad \frac{4}{(n+1)^2}<|a_n-a_{n+1}|^2<\frac{4}{n^2}. $$ Thus $$ \sum |a_n-a_{n+1}|^2<\infty\quad \text{while}\quad \sum |a_n-a_{n+1}|=\infty $$