$x_{n+1}\le x_n+\frac{1}{n^2}$ for all $n\ge 1$ then did $(x_n)$ converges?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $L$.