If $\{a_n\}$ is bounded above and $a_{n+1} - a_{n} > -\frac{1}{n^{2}}$, then is $\{a_n\}$ convergent? Suppose that a sequence $\{a_n\}$ is bounded above and if satisfies the condition $a_{n+1} - a_{n} > \frac{-1}{n^{2}},$ where $n$ is a natural number. Then $\{a_n\}$ is convergent. $($true or false$)?$
$a_{n+1} - a_{n} > \frac{-1}{n^{2}}$
$a_n - a_{n+1} < \frac{1}{n^{2}}$
So, this sequence is monotonically decreasing.
And $$\lim_{n \to \infty} (a_n - a_{n-1})= 0$$
How to proceed to see, if the sequence converges$?$
 A: $$\frac{1}{n^2} = b_n-b_{n-1}, \ n\geq2, \ b_1 = 1 $$
then
$$b_n = \sum_{k=1}^n \frac{1}{k^2} $$
Then the following sequence is convergent, because it's increasing and bounded:
$$x_n = b_n+a_n $$
Hence $a_n = x_n-b_n$ is also convergent.
A: First of all $a_{n}-a_{n+1}<\frac{1}{n^{2}}$ doesn't mean the sequence is monotonically decreasing, on the contrary means that tends to be monotonically increasing.
For example the sequence $a_{n}=n$ satisfy the inequality.
In fact $a_{n}-a_{n+1}< 0$ means monotonically increasing and $a_{n}-a_{n+1}>0$ decreasing.
The proof of your question is that: the series $1/n^{2}$ converges, so $\forall\varepsilon>0\,\exists n_{\varepsilon}$ tc$\forall N>n_{\varepsilon}$ $\sum_{n=N}^{\infty}1/n^{2}<\varepsilon$.
Because the sequence is upper bounded we can call $M:=\limsup_{n\to\infty}a_{n}$, and we have a subsequence $a_{n_{k}}$ which converges to M. So $\forall\varepsilon>0$ $\exists k_{\varepsilon}$ tc $|M-a_{n_{k_{\varepsilon}}}|<\varepsilon$.
From the hypotesis we know that if $n>n_{k}$ then $a_{n_{k}}-a_{n}<\sum_{i=n_{k}}^{n}1/i^{2}<\sum_{i=n_{k}}^{\infty}1/i^{2}$, so $a_{n}>a_{n_{k}}-\sum_{i=n_{k}}^{infty}1/i^{2}$.
It follows that for $k>\max\{k_{\varepsilon},n_{\varepsilon}\}$ $a_{n}>M-2\varepsilon$ $\forall n>n_k$.
On the other hand let $M_{k}:=\sup_{n>n_k}\{a_{n}\}$, so we have that obviously $a_{n}<M_{k}$ $\forall n>k$, and it's clear because of the definition that $M_{k}\to M$, and it does it decreasingly.
So for $k>m_{\varepsilon}$ we have $M_{k}<M+\varepsilon$.
So for $k>\max\{k_{\varepsilon},n_{\varepsilon}, m_{\varepsilon}\}$ and for $n>n_{k}$ we have $M+\varepsilon>a_{n}>M-2\varepsilon$, that means $a_n\to M$.
