Let $x_n$ be a sequence of non-negative real numbers such that $\forall n, x_{n+1}\leq x_n+ \frac{1}{n^2}$
Prove that $x_n$ converges.
The problem is trivial whenever $x_{n}$ is an increasing sequence.
In the general case I managed to prove that $(x_n)$ is bounded. It remains to prove that $(x_n)$ has only one accumulation point, but that seems difficult.
Any hint is welcome.
For $m \gt n \gt 1$ you have $x_m-x_n \le \dfrac{1}{n-1}$
so any two accumulation points must be arbitrarily close (use $m$ for a subsequence approaching the higher accumulation point and $n$ for a subsequence approaching the lower accumulation point) and so any accumulation points must be equal.
Because the series $\sum_{n=1}^\infty\dfrac1{n^2}$ converges, to each $\varepsilon>0$ there is an $N(\varepsilon)$ such that $\sum_{n>N}n^{-2}<\varepsilon$.
If $A,B$ were two distinct accumulation points, say with $A<B$, let $\varepsilon=(B-A)/3$. After having visited an $\varepsilon$-neighborhood of $A$ after $n>N(\varepsilon)$ you can never get near $B$ again.
Hint:
Since $(x_n)$ is bounded, there is a convergent subsequence $x_{n_k}\to y\ge0$. Show that $\forall \epsilon>0$, $\exists N>0$, $\forall n>N$, $|x_n-y|<\epsilon$.
(Note that $x_n$ is between some $x_{\displaystyle{n_k}}$ and $x_{\displaystyle{n_{k+1}}}$ , and that $\sum_{n=1}^\infty\dfrac{1}{n^2}<\infty.$)
We will show the sequence is Cauchy. Perhaps the steps in this argument will help you understand some of the results others stated without support.
As in the case of more straightforward limit proofs, we first do some scratch work. So let $m, n \in \mathbb{N}$ be given, and assume $m = n + k$ for some $k > 0$. We have to bound the difference $|x_{n+k} - x_n|$. By the triangle inequality,
\begin{align*} |x_{n+k} - x_n| &\leq |x_{n+k} - x_{n+k-1}| + |x_{n+k-1} - x_{n+k-2}| + \cdots + |x_{n+2} - x_{n+1}| + |x_{n+1} - x_n| \\ &\leq \frac{1}{(n+k-1)^2} + \frac{1}{(n+k-2)^2} + \cdots + \frac{1}{(n+1)^2} + \frac{1}{n^2} \\ &< \sum_{i=n}^\infty \frac{1}{i^2}. \end{align*}
Knowing that the summation converges ($p$-series with $p = 2 > 1$), we have that for any $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n > N$, $$\sum_{i=n}^\infty \frac{1}{i^2} = \left|\sum_{i=n}^\infty \frac{1}{i^2}\right| = \left|\sum_{i=1}^n \frac{1}{i^2} - S\right| < \varepsilon.$$ In other words, we can make the summation small. Now for the proof.
Let $\varepsilon > 0$ be given, and choose $N \in \mathbb{N}$ such that $\sum_{i=n}^\infty i^{-2} < \varepsilon$ for all $n > N$. Then whenever $m > n > N$, $$|x_m - x_n| < \sum_{i=n}^\infty \frac{1}{i^2} < \varepsilon.$$
Aside: this technique of proof mirrors that of the Contraction Mapping Theorem. Even if you haven't reached this part of your studies yet, I recommend reading the theorem and proving it on your own following this proof as a guideline.
