How can I tell that the sequence $a_n=\frac {\ln(n)} {n}$ converges and to what it converges?
Emily Wong 
I need to "study the limit behavior" and compute the limit if it exists.
This is what I have done so far. In order to study the limit behavior I tried to first check the monotonicity and boundedness of the sequence. The sequence was found to not be monotonic for all n. Since that failed, I tried to attempt to prove that the sequence was Cauchy, and if it was, that would lead me to have completed the "study the limit behavior" part. Here is my attempt at the cauchy part:
For every ϵ>$0$ there exists an $N$ such that $m,n>N$ implies |$a_n - a_m|<ϵ$ for $n\geq m$ $$|a_n-a_m|=\bigg|\frac {\ln(n)} {n}-\frac {\ln(m)} {m}\bigg|\leq \bigg|\frac {\ln(n)} {n}\bigg|+\bigg|\frac {\ln(m)} {m}\bigg|\Leftrightarrow \frac {\ln(n)}{n}+\frac{\ln(m)}{m}<ϵ$$ $$\frac {\ln(n)}{n}+\frac{\ln(m)}{m}<ϵ \Leftrightarrow \frac {\ln(n)}{n}<ϵ- \frac {\ln(m)}{m}$$ (Now I get stuck. I dont know if what I did so far is even correct, and if it is I don't know where to go from here)
Even though I got stuck at the cauchy part I went on to compute the limit. $$\lim a_n=\lim\frac{\ln(n)}{n}=\lim\frac{1}{n}\bigg(\ln(n)\bigg)=\lim \ln(n)^{\frac{1}{n}}$$ So, $\lim {e^{a_n}}=\lim e^{\ln(n)^{\frac{1}{n}}}=\lim n^{\frac {1}{n}}=1$ ( I proved that using the epsilon definition)
Therefore because $\lim e^{a_n}=1\ \lim \ln(e^{a_n})= \lim a_n=\ln(1)=0$
Any help (on the cauchy part especially)? Thanks in advance.
$\endgroup$ 23 Answers
$\begingroup$You made a mistake
$$\left|\frac{\log(m)}{m}-\frac{\log(n)}{n}\right| \leq \left|\frac{\log(m)}{m}\right| - \left|\frac{\log(n)}{n}\right|$$
is wrong, the rhs could be smaller zero the lhs not.
The function is monotone for $n$ large enough (for $n>e$), so just prove monoticity for $n>3$.
Consider the continuous version:
$$f(x) = \frac{\ln x}{x}$$
Computing the derivative:
$$ f'(x) = \frac{1 - \ln x}{x^2} $$
When $x > e$, this derivative is negative and so it is a decreasing function for $x > e$. On the other hand, it's clear that this function is also bounded above $0$ and so the sequence $a_n$ is bounded above $0$ and is decreasing for $n > 2$. Monotone convergence theorem says that $a_n$ must then converge.
To find what it converges to, we could use L`Hopital's rule:
$$\lim_{x\rightarrow \infty} f(x) = \lim_{x\rightarrow \infty} \frac{1}{x} = 0$$
$\endgroup$ 1 $\begingroup$You can use the fact that $n^{1/n}\rightarrow 1$ as $n\rightarrow \infty$.
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