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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 $\lim_{n\rightarrow \infty} c_n =0$

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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.

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  • $\begingroup$ Nice variation on the problem :) $\endgroup$
    – user459879
    Nov 29 '18 at 23:36
  • $\begingroup$ You are welcome! $\endgroup$ Nov 30 '18 at 1:20

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