$x_{n+1}\le x_n+\frac{1}{n^2}$ for all $n\ge 1$ then did $(x_n)$ converges? If $(x_n)$ is a sequence of nonnegative real numbers such that $x_{n+1}\le x_n+\frac{1}{n^2}$ for all $n\ge 1$ then did $(x_n)$ converges?
Can someone help me please?
 A: Let 
$$s_n:=\sum_{k=1}^{n-1} \frac{1}{k^2} \,.$$
Then $S_n$ is convergent, hence bounded.
Note that $S_{n+1}-S_n=\frac{1}{n^2}$. Thus
$$x_{n+1}\le x_n+S_{n+1}-S_n \Rightarrow x_{n+1}-S_{n+1}\le x_n-S_n$$
Hence $s_n-S_n$ is decreasing. As its is the difference of two bounded sequences, it is also bounded (we only care about a lower bound, and $-\frac{\pi^2}{6}$ is such a lower bound, but it doesn't matter.)
As $x_n-S_n$ is monotonic and bounded, it is convergent.
Then
$$x_n=(x_n-S_n)+S_n$$
is the sum of two convergent sequences, thus convergent.
A: Yeah it does. 
Note that we have $x_{n+1}-x_n\leq \frac{1}{n^2}$. Hence $$\mid x_{n}-x_m\mid \le \sum_{i=m}^{n-1}\frac{1}{i^2}$$
Its thus a Cauchy sequence and hence converges.
A: Since 
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2) = \frac{\pi^2}{6} < \infty,$$
$x_{n}$ is bounded from above by $x_1 + \frac{\pi^2}{6}$. Since $x_n$ is also non-negative, it is a bounded sequence and both hence its lim sup and lim inf exists. Let $L$ be the lim inf of $x_n$.
For any $\epsilon > 0$, pick a $N$ such that $\displaystyle \sum_{n=N}^\infty \frac{1}{n^2} < \frac{\epsilon}{2}$.
By definition of $L$, there is a $M > N$ such that $x_M < L + \frac{\epsilon}{2}$. For any $n > M$, we have
$$x_n 
\;\;<\;\; x_M + \sum_{k=M}^{n-1} \frac{1}{k^2} 
\;\;<\;\; L + \frac{\epsilon}{2} + \sum_{k=M}^{\infty}\frac{1}{k^2} 
\;\;<\;\; L + \epsilon
$$
This implies 
$$\limsup_{n\to\infty} x_n \le L + \epsilon$$
Since $\epsilon$ is arbitrary, we get 
$$\limsup_{n\to\infty} x_n \le L = \liminf_{n\to\infty} x_n
\implies 
\limsup_{n\to\infty} x_n = \liminf_{n\to\infty} x_n = L
$$
i.e. the limit exists and equal to $L$.
