I am not sure if I'm doing it correctly. The two sequences are as follows:
(1). Let $\{a_n\}$ be a sequence of real numbers such that $|a_{n+1}-a_n|<\frac{1}{n}$ for every $n\in \mathbb{N}$.
(2). Let $\{b_n\}$ be a sequence of real numbers such that $|b_{n+1}-b_n|<\frac{1}{n^2}$ for every $n\in \mathbb{N}$.
For (2), I argued in this way:
$$\sum_{n=1}^{\infty}|b_{n+1}-b_n|<\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty.$$ Thus, $\lim_{n\rightarrow \infty}|b_{n+1}-b_n|=0$, which means that $b_n$ converges. Therefore, $b_n$ is Cauchy.
But for (1), I found the similar argument doesn't work so I tried to compute $|a_{m}-a_n|$ for $m>n$. And I got something like $|a_{m}-a_n|<\frac{1}{m-1}+\frac{1}{m-2}+\cdots+\frac{1}{n}$. So I think $\{a_n\}$ is not a Cauchy sequence, but I don't know how to proceed.
Do you have any suggestion and can you help me to verify whether I am correct?
For (1), consider $$a_n=\frac12\left(1+\frac12+\frac13+\cdots+\frac1n\right).$$ This sequence diverges (despite $\lim_{n\to\infty}|a_{n+1}-a_n|=0$).
For (2), can you prove that $$|b_n-b_m|<\sum_{k=\min(m,n)}\frac1{k^2}?$$