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.

  • $\begingroup$ It was Dominic who edited it for you, not me. It not that you're supposed to edit it like this, but that's how $\LaTeX$ is supposed to be used. $\endgroup$ – Git Gud Mar 15 '13 at 7:20
  • $\begingroup$ Oh okay, thanks for the info. $\endgroup$ – user66807 Mar 15 '13 at 7:23

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 $n>3$.

  • $\begingroup$ I tried that but doesnt the sequence have to be monotone for all n and because the sequence isn't monotone for n<3 I can't use the monotonicity of the sequence. $\endgroup$ – user66807 Mar 15 '13 at 7:03
  • $\begingroup$ as i said go for $n>3$ there it is monotone, at what happens below $3$ doesn't matter for the limit at all $\endgroup$ – Dominic Michaelis Mar 15 '13 at 7:07
  • $\begingroup$ That is what I did first, but my book says that a sequence is decreasing if $s_n>_=s_{n+1} for all n$ and a decreasing sequence will be called monotone. Also, what I did was the triangle inequality that says $|a+b|<_=|a|+|b|$ Does that not work? $\endgroup$ – user66807 Mar 15 '13 at 7:13
  • $\begingroup$ the triangle inequality works but it gives you $|a-b|\geq |a|-|b|$, which is not helpful. You could use $|a-b|\leq |a|+|b|$ with that one you can go. Yeah your original series is not monotone for all $n$, but we are interested in the limit, so the first few values don't matter at all. $\endgroup$ – Dominic Michaelis Mar 15 '13 at 7:17
  • $\begingroup$ Oh okay, I understand what you are saying now, but since the problem says to study the behavior of the limit don't I have to know the sequence converges before I can talk about the limit? And for the inequality part I understand what you are saying, do I just edit the question? $\endgroup$ – user66807 Mar 15 '13 at 7:21

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:

$$\lim_{x\rightarrow \infty} f(x) = \lim_{x\rightarrow \infty} \frac{1}{x} = 0$$

  • $\begingroup$ Although I have learned L`Hopital's rule and derivatives, because we haven't gotten to it in this class I don'think I'm allowed to use them. $\endgroup$ – user66807 Mar 15 '13 at 7:25

You can use the fact that $n^{1/n}\rightarrow 1$ as $n\rightarrow \infty$.


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