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Let $f: [1,\infty) \rightarrow R$ be a function. Suppose that f is increasing, prove that if $\lim_{x \rightarrow \infty} f(x)$ exists then the sequence $\{f(n)\}_{n=1}^{\infty}$ is convergent.

So by definition of limit of function at infinity, let $\lim_{x \rightarrow \infty} f(x)=L$, then for any $\epsilon>0$, there exists $M \in R$ such that for $x \in [1,\infty)$, $x>M$, we have $|f(x)-L|<\epsilon$. I want to connect this to definition of limit of sequence: for any $\epsilon>0$, there exists $N \in \mathbf{N}$ such that $n>N$ implies $|f(n)-L|<\epsilon$.

It seems pretty easy, can I just let $N=M$, and $x=n$ (I have one concern though: $M \in R$, but N needs to be in $\mathbf{N}$.


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So strange. I am pretty sure this exact same question was asked before, with a lot of comments. I am unable to find it though. – Aryabhata Apr 13 '11 at 21:53
@Moron: Yes, same déjà-vu here, but I'm pretty sure that last time it was an if and only if statement... – t.b. Apr 13 '11 at 21:59
It was the other direction of the statement, right? – GWu Apr 13 '11 at 22:07
@GWu: Well, the question was to prove that for increasing $f$ convergence of $f(x)$ is equivalent to convergence of $f(n)$. It provided an ok argument for the direction the question here asks about and said "read the argument backwards" for the (slightly more) interesting direction. – t.b. Apr 13 '11 at 22:11
up vote 3 down vote accepted

You don't need monotonicity of $f$. And yes, you have the right proof, except that you just need to take $N=\lceil M \rceil$.

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The hypotheses imply that $f$ is bounded "at infinity". Hence the sequence $(f(n))$ will be increasing and bounded and so converges (to its sup).

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I think GWu's answer is better. As he points out, "increasing" is simply not needed here, and your implicit appeal to completeness seems like overkill. For instance, the result also holds for functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$, right? – Pete L. Clark Apr 14 '11 at 6:18
@Pete, you're right of course. – lhf Apr 14 '11 at 10:08

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