Will it be a Cauchy sequence? 
Let $<x_n>$ be a sequence satisfying
  $|x_{n+1}-x_n|\le \frac{1}{n^2}$
  Will it be a Cauchy sequence?

 A: We want to show that given any $\epsilon \gt 0$, there is an $N$ such that if $N\le m\lt n$ then $|a_n-a_m|\lt \epsilon$.  Note that
$$a_n-a_m=(a_{m+1}-a_m)+(a_{m+2}-a_{m+1})+\cdots+(a_n-a_{n-1}).$$
Taking absolute values, and using the Triangle Inequality, we get
$$|a_n-a_m|\le |a_{m+1}-a_m|+|a_{m+2}-a_{m+1}|+\cdots+|a_n-a_{n-1}|.$$
The right-hand side is non-negative and less than or equal to
$$\frac{1}{m^2}+\frac{1}{(m+1)^2}+\cdots+\frac{1}{(n-1)^2}.$$
Since $\sum_1^\infty \frac{1}{k^2}$ converges, by choosing $m$ large enough, we can make the tail $\lt \epsilon$. If you want to be fully explicit, note that
$$\frac{1}{m^2}+\frac{1}{(m+1)^2}+\frac{1}{(m+2)^2}+\cdots \lt \frac{1}{(m-1)m}+\frac{1}{(m)(m+1)}+\frac{1}{(m+1)(m+2)}+\cdots,$$
and the series on the right is a telescoping series with sum $\frac{1}{m-1}$.
If omitted detail makes the solution obscure, please leave a message.
Remark: The same is true if $\frac{1}{n^2}$ is replaced by $\frac{1}{n^p}$, where $p$ is any real number $\gt 1$. But we cannot replace $\frac{1}{n^2}$ by $\frac{1}{n}$. That is because the harmonic series diverges.
A: Let $m>n$ then we have
$$|x_m-x_n|\leq \sum_{k=n}^{m-1}|x_{k+1}-x_k|\leq \sum_{k=n}^{m-1}\frac{1}{k^2}\to_{m,n\to\infty}0$$
since the series $\displaystyle\sum_k \frac{1}{k^2}$ is convergent so we conclude that $(x_n)$ is a Cauchy sequence.
A: As a fairly obvious generalization,
if
$|x_{n+1}-x_n|\le f(n)$
and
$\sum_{n > 0} f(n)$
converges,
then $(x_n)$ is Cauchy.
Going the other way,
if 
$\sum_{n > 0} f(n)$
diverges,
then we can find a sequence
$(x_n)$
such that
$|x_{n+1}-x_n|\le f(n)$
and
$(x_n)$ diverges.
