# Let $(a_n), (b_n),$ be bounded, then prove that $c_n$ converges and give its value.

Could I get some feedback on the following proof, I feel becoming a good mathematician is through constant feedback and improvement of your work, I tried to make it short, but well-readable.

We are given that $$a_n$$ and $$b_n$$ are bounded sequences, and also: $$(n-1)a_n \leq n^2 c_n \leq (n+1)b_n$$ Prove that $$c_n$$ converges and give the value of the limit.

First of all, we know that these sequences are bounded, so surely we have that: $$L \geq a_n \land b_n \leq U$$ For some lower bound $$L$$ and some upper bound $$U$$ in $$\mathbb{R}$$. We now apply this: $$(n-1) L \leq (n-1)a_n \leq n^2 c_n \leq (n+1)b_n \leq (n+1)U$$ $$\frac{L}{n}-\frac{L}{n^2}=\frac{(n-1) L}{n^2} \leq c_n\leq \frac{(n+1) U}{n^2} =\frac{U}{n}+\frac{U}{n^2}$$ Both these sequence converge to zero, so we have that in the limit: $$0 \leq \lim_{n\rightarrow \infty} c_n \leq 0$$ By the squeeze theorem we have that $$\lim_{n\rightarrow \infty} c_n =0$$

Absolutely fine proof. You may want to find if the given condition can be relaxed. For example, if $$n^2$$ is replaced by $$n^\alpha$$ in the above proof, which values would work? $$n = n^1$$ would not work : try to find a counterexample.