# Boundedness of a Sequence

Question : How do we prove that the sequence (1+(1/n))^n is bounded?

My thoughts on this question : We know that, to prove a sequence is bounded, we should prove that it is both bounded above and bounded below.

We can see that all the terms in the sequence is greater than 0. Which means the sequence is bounded below.
But gow do we prove that it is bounded above. Does taking the limit of the general term(say nth term) as n tends to infinity always gives us the highest possible term of the sequence, which will give us the opportunity to find an upper bound of the sequence?

Thanks!

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1. Hmmm. By the Binomial Theorem,
(1 + 1/n)^n
= 1 + n*1/n + n(n-1)/2!n^2 + n(n-1)(n-2)/3!n^2 + ...
< 1 + 1 + 1/2! + 1/3! + ...
which you can show converges

I expect google can turn up various more direct proofs.

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oobleck
2. I Googled this and find that proof. But can you please let me know whether we can take the highest value of a monotonically increasing sequence by taking the limit of its general term(say nth term), when n goes to infinity?
Can we consider this value(if finite), as an upper bound of this sequence?

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