I'm really stuck with my homework in mathematical analysis. I would be very glad, if you could give me some ideas/solutions or tips, how to get this done. The task is as follows:
Let $(x_{n})$ be a bounded sequence and let M $\in$ $\mathbb{R}$. Prove, that
$\limsup\limits_{ n\to\infty} x_{n} $ $\leq$ M
if and only if for every $\varepsilon > 0 $, there exists an N $\in$ $\mathbb{N}$ so, that
$x_{n} < M + \varepsilon$ for every $n \geq N$.
To give you some frame I will solve one side, and will leave the other side to you.
Denote: $$s_k:=\sup\{x_n\mid n\geq k\}$$ Then $(s_k)$ is a decreasing sequence, and by definition: $$\limsup_{n\to\infty}x_n=\lim_{k\to\infty}s_k\tag1$$
If $x_n<M+\epsilon$ for every $n\geq N$ then $M+\epsilon$ is an upper bound for the set $\{x_n\mid n\geq N\}$. Consequently the least upper bound of this set will not exceed $M+\epsilon$:$$s_N\leq M+\epsilon$$ Since $(s_k)$ is decreasing we have:$$k\geq N\implies s_k\leq M+\epsilon$$ so that: $$\lim_{k\to\infty}s_k\leq M+\epsilon$$
This for every $\epsilon>0$ so finally, and applying $(1)$ we find:$$\limsup_{n\to\infty}x_n=\lim_{k\to\infty}s_k\leq M$$