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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?
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.
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.
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 $$s_n=\sum_{k=0}^nx_n\le nx_0+\sum_{k=1}^{n-1}\frac{n-k}{k^2}$$
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 $x_n\to L$.
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 $-\infty$.
