Let ${x_n}$ be a sequence in $R$ such that $|x_{n+1} - x_n|< \frac{1}{n^2}$ for all $n \in N$.
Show that the sequence is convergent.
If it were $|x_{n+1} - x_n|= \frac{1}{n^2}$, could take help of the fact $\sum \frac{1}{n^2}$ is convergent and of triangular inequality.
But what to do here? Please help. Thanks in advance.
We know that $\sum_{k\geq1}\frac{1}{k^{2}} $ is convergent, then for all $\epsilon>0 $ exists some $N\in\mathbb{N} $ such that for all $m,n>N $ we have $$\sum_{k=n}^{m}\frac{1}{k^{2}}<\epsilon $$ so $$\epsilon>\sum_{k=n}^{m}\frac{1}{k^{2}}>\left|x_{m+1}-x_{m}\right|+\dots+\left|x_{n+1}-x_{n}\right|\geq\left|x_{m+1}-x_{n}\right| $$ so it is a Cauchy sequence and so it is convergent.
We can also rephrase the problem so that we study series rather than sequences.
If we denote $a_1=x_1$ and $a_n=x_n-x_{n-1}$ for $n\ge2$, then $$x_n = \sum_{k=1}^n a_k.$$ (It is a telescoping series.)
The question whether the sequence $x_n$ is convergent is equivalent to the question whether the series $\sum a_n$ is convergent. (By definition, convergence of a series is equivalent to convergence of the partial sums.)
Since $$\sum_{k=1}^\infty |a_k| \le \sum_{k=1}^\infty \frac1{k^2} < +\infty,$$ the series $\sum a_n$ is absolutely convergent and therefore convergent.
So the sequence $x_n$ is convergent, too.