Sunday, March 31, 2019

calculus - Show that the sequence $lim_{nrightarrowinfty} log(n)$ does not converge




In order to show that the sequence $$\lim_{n\rightarrow\infty} \log(n)$$ does not converge I have to show:



there exists $N \geq 0$ s.t $$\left|a_{n} - L \right| \ > \epsilon$$ for each $N$ and for all choices of $L$.



Here is where my issue lies. I'm trying to find a relationship to apply. I asked for some help from my prof and he suggested $$ \log(n) > M \implies n > e^{M} $$



Now what comes to my mind from this is to somehow make this $M$ my $\epsilon$, but I don't see where to carry through the comparison..


Answer



It's enough to push the values of the sequence off to infinity, this is the purpose of the hint.




I.e. Given any positive $M$, you ought to be able to find an $N$ large enough such that for any $n>N$
$$
\log(n)>M
$$
your professor told you how to find this $N$, namely
$$
\log(n)>M\iff n>e^M
$$
so picking $n=\lceil e^M \rceil$ will do. Since choice of $M$ was arbitrary, you are done.



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