3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.