Prove that $\lim_{n\rightarrow\infty}\frac{exp(r)}{r^n}=\infty$ from the definition.
The definition in my text says: For a sequence $(S_n)$, we write $limS_n=\infty$ provided for each $M>0$ there is a number N such that $n>N$ implies that $S_n>M$.
So I need to consider an arbitrary $M>0$ and show that there exists $N$ [which will depend on $M$] such that $n>M$ implies $\frac{exp(r)}{r^n}>M$
To see how big N must be we normally solve for n in the inequality.
So far I have only had to do limit definitions for equalities and inequalities that have just one variable but in this case there is an $r$ and $n$ and even if I knew which one to solve for (I think $r$ in this case) I'm not sure how to isolate it since the $r$ is in both the numerator and denominator.
So my question is how can I choose my $n>M$ for this case?
Note: I know how to show the limit is infinite using limits and l'Hopital's rule but I'm not allowed to use this as my formal proof I must use the definition.
Any help would be greatly appreciated.
Thank you,
You should solve for n: $$\frac{e^r}{r^n} = \frac{e^r}{e^{(\ln{(r^n)})}} = \frac{e^r}{e^{(n \cdot\ln{r})}} = e^r\cdot e^{n\cdot(-\ln{r})}$$ If you want a tight bound you can solve the following inequality for n: $$e^r\cdot e^{n\cdot(-\ln{r})} > M$$ $$e^{n\cdot(-\ln{r})} > \frac{M}{e^r}$$ $$n \cdot (-\ln{r}) > \ln{\left(\frac{M}{e^r}\right)}$$ $$n > \frac{\ln{\left(\frac{M}{\exp{(r)}}\right)}}{-\ln{r}}$$ $$n > \frac{\ln{\left(\frac{M}{\exp{(r)}}\right)}}{\ln{\left(\frac{1}{r}\right)}}$$ $$n > \ln{\left(\frac{M}{e^r}-\frac{1}{r}\right)}$$