Prove: If $\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=L$ and ${a}_{n}>a$ for all n then $L\ge a$

Freddy Friedman
2022-07-17
Answered

Prove: If $\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=L$ and ${a}_{n}>a$ for all n then $L\ge a$

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eishale2n

Answered 2022-07-18
Author has **15** answers

Suppose that $L<a$. We put $\epsilon =a-L>0$. Since $\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=L$, there exists ${N}_{0}\in \mathbb{N}$ such that

$|{a}_{n}-L|<\epsilon \phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge {N}_{0}.$

Then ${a}_{n}-L<a-L$ for all $n\ge {N}_{0}$, or ${a}_{n}<L$. This is contradict to the assumption ${a}_{n}>a$ for all n. Hence $L\ge a$. In the above argument, we have seen that we only need ${a}_{n}>L$ for sufficiently large n.

Moreover, in the general case we do not have $L>a$. Indeed, we observe that although ${a}_{n}=\frac{1}{n}>0$ for all n but $L=\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}=0$

$|{a}_{n}-L|<\epsilon \phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge {N}_{0}.$

Then ${a}_{n}-L<a-L$ for all $n\ge {N}_{0}$, or ${a}_{n}<L$. This is contradict to the assumption ${a}_{n}>a$ for all n. Hence $L\ge a$. In the above argument, we have seen that we only need ${a}_{n}>L$ for sufficiently large n.

Moreover, in the general case we do not have $L>a$. Indeed, we observe that although ${a}_{n}=\frac{1}{n}>0$ for all n but $L=\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}=0$

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