How to prove this sequence converges Here is a problem in analysis:

Suppose $x_n\geq0$ and for all $n$, there is
  $$ 
x_{n+1}\leq x_n+\dfrac1{n^2}
$$
  Prove that $x_n$ converges.

My approach: it is easy to prove $x_m-x_n\leq \epsilon$ using telescope series. But in order to prove it is a Cauchy sequence, it has to be proved that $x_n-x_m\leq \epsilon$ too. I am not sure how to prove the second step.
 A: Note that $\displaystyle x_{n+k}\leq x_n+\frac{1}{n^2}+\cdots+\frac{1}{(n+k-1)^2}\leq x_n+\frac{1}{n^2}+\cdots$. We have
\begin{eqnarray}
\varlimsup_{k\to\infty}x_{n+k}\leq x_n+\frac{1}{n^2}+\cdots
\end{eqnarray}
Let $\displaystyle\varepsilon(n)=\frac{1}{n^2}+\cdots$. We have $\displaystyle\lim_{n\to\infty}\varepsilon(n) =0$ as the series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}$ is known to converge. Now taking liminf on the RHS we have (note that it makes sense to take liminf because $x_n$ is assumed to be nonnegative)
\begin{eqnarray}
\varlimsup_{n\to\infty}x_n=\varlimsup_{k\to\infty}x_{n+k}\leq\varliminf_{n\to\infty}(x_n+\varepsilon(n))=\varliminf_{n\to\infty}x_n+\lim_{n\to\infty}\varepsilon(n)=\varliminf_{n\to\infty}x_n
\end{eqnarray}
So the limsup and liminf of the sequence coincide and it converges.
A: Hint:
$$x_1\le x_0+1$$
$$x_2\le x_1+\frac14\le x_0+1+\frac14$$
$$x_3\le x_2+\frac19\le x_0+1+\frac14+\frac19$$
$$\cdots$$
$$x_n\le x_0+\sum_{k=1}^n\frac1{k^2}.$$
A: Since $\sum\limits_{n=1}^{\infty}\dfrac1{n^2}$ converges, by Cauchy's Criterion
$$
\forall \epsilon>0, \exists N>0, \forall m>n>N,\:\:\sum\limits_{k=n}^{m}\dfrac1{k^2}<\epsilon
$$
By the telescopic nature of the series, for all $m>n>N,\,$ there is
$$
x_m-x_n<\sum\limits_{k=n}^{m-1}\dfrac1{k^2}<\epsilon \hspace{5 mm} \text{or} \hspace{5 mm} x_m<x_n+\epsilon
$$
Fix $n$, take limsup as $m\to\infty$
$$
\varlimsup\limits_{m\to\infty}x_m\leqslant x_n+\epsilon
$$
Since the above holds for all $n>N$, take liminf as $n\to\infty$ and we get
$$
\varlimsup\limits_{m\to\infty}x_m\leqslant \varliminf\limits_{n\to\infty}(x_n+\epsilon)=\varliminf\limits_{n\to\infty}x_n+\epsilon
$$
Since $\epsilon$ is arbitrary, we have
$$
\varlimsup\limits_{m\to\infty}x_m\leqslant \varliminf\limits_{n\to\infty}x_n \hspace{5 mm} \text{or} \hspace{5 mm} \varlimsup\limits_{n\to\infty}x_n= \varliminf\limits_{n\to\infty}x_n 
$$
So $\lim\limits_{n\to\infty}x_n$ exists.
