Existence of limit for a given sequence: $x_{n+1} \le x_n + 1/{n^2}$ Let $x_n$ be a sequence in $\mathbb{R}$ such that
$$x_{n+1} \le x_n + \frac{1}{n^2}$$
Prove that $\lim x_n$ exists (it can be a real number or infinite).
I've tried to prove it using the delta-epsilon definition, limit superior and inferior, Cauchy Criterion for convergence of sequences, all kinds of convergence criteria for sequences... but nothing seems to work.
Can you help me?
 A: Let's define a new sequence $y_n = x_n - \displaystyle\sum_{m = 1}^{n-1}\dfrac{1}{m^2}$. 
Then, $y_{n+1} = x_{n+1} - \displaystyle\sum_{m = 1}^{n}\dfrac{1}{m^2} = x_{n+1} - \dfrac{1}{n^2} - \displaystyle\sum_{m = 1}^{n-1}\dfrac{1}{m^2} \le x_n - \displaystyle\sum_{m = 1}^{n-1}\dfrac{1}{m^2} = y_n$. 
Since $y_{n+1} \le y_n$ for all $n$, we have that $y_n$ is strictly decreasing. 
Therefore, either $\displaystyle\lim_{n \to \infty}y_n = L$ or $\displaystyle\lim_{n \to \infty}y_n = -\infty$. Can you finish from here?
I'm assuming that you can prove that $\displaystyle\sum_{m = 1}^{\infty}\dfrac{1}{m^2}$ converges.
A: Assuming that the sequence $\{x_n\}_{n\in\mathbb{N}}$ has two distinct accumulation points $A$ and $B$, then $x_n$ must belong to an $\epsilon$-neighbourhood of $A$ infinitely often and do the same of an $\epsilon$-neighbourhood of $B$. However, the series $\sum_{n=1}^{+\infty}$ is convergent, so there exists a $N$ such that $\sum_{n\geq N}\frac{1}{n^2}<\frac{|A-B|}{2}$. 
For any $n\geq N$, the sequence is so trapped in a neighbourhood of $A$ or in a neighbourhood of $B$, contradicting the existence of two different accumulation points.  
A: For $n\geq 2$, we have $\displaystyle \frac{1}{n^2}\leq \frac{1}{n(n-1)}$
hence
$$x_{n+1}\leq x_n+\frac{1}{n(n-1)}=x_n+\frac{1}{n-1}-\frac{1}{n}$$
Put now
 $\displaystyle u_n=x_n+\frac{1}{n-1}$ .For $n\geq 2$. We have 
$\displaystyle u_{n+1}\leq u_n$. Now there is two
possibilities: $u_n\to -\infty$, and $x_n\to -\infty$, or $u_n\to L\in \mathbb{R}$, and $x_n\to L$. 
A: Let's define partial sums as:
$$s_n=\sum_{k=0}^nx_n=x_0+x_1+\cdots+x_n$$
As:
$$x_{n+1}\le x_n+\frac1{n^2}$$
So
$$s_n=x_0+x_1+\cdots+x_n\le\left(x_0+x_1+\cdots+x_{n-1}\right)+\left[x_{n-1}+\frac1{(n-1)^2}\right]$$
Or
$$s_n=\sum_{k=0}^nx_n\le nx_0+\sum_{k=1}^{n-1}\frac{n-k}{k^2}$$ 
A: The positive part of the series $\sum_n(x_{n+1}-x_n)$ clearly converges,
$$
\sum_n(x_{n+1}-x_n)_+\leq\sum_{n}1/n^2.
$$
It follows that
$$
\lim_{n\rightarrow\infty}x_n=\sum_n(x_{n+1}-x_n)=\sum_n(x_{n+1}-x_n)_+-\sum_n(x_{n+1}-x_n)_-
$$ exists, possibly with the value $-\infty$.
