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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 $\mathbf{N}$.

Thanks!

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    @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.2011-04-13

2 Answers 2

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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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    @Pete, you're right of course.2011-04-14