Does the sequence ln(n)²/n converge to 0?

[Hall]

I'm simply amazed at the apparent impossibility of isolating $n$ in
$$
\big| \frac{ 2 ln(n)}{n}\big| \lt \varepsilon
$$
and the impossibility of Squeezing $\frac{2 ln (n)}{n}$ properly. There is no function that I can think of which does the job.

I do not know if the following proof is rigorous, but I present it still:
$$
\begin{flalign*}
a_n = \frac{ 2 ln(n) }{n} \\
A_n = e^{a_n} = (e^{ln(n)})^{2/n} = (n^{1/n})^2\\
\lim A_n = \lim n^{1/n} \cdot \lim n^{1/n} = 1\\
\lim e^{a_n} = 1 \\
\textrm{As}~e^x~\textrm{is a continuous function, we have} \\
\lim a_n = a_0 \implies \lim e^{a_n} = e^{a_0} = 1 \implies a_0 = 0 \\
\textrm{Thus,} ~~\lim a_n = 0\\
\end{flalign*}
$$

 

[chwala]

This argument doesn't really work. Suppose instead we assume $n=1$ instead of $n=3$. The rest of the argument would remain unchanged, but it would be wrong to conclude the sequence is decreasing after $n=1$.

Your assumption is that the sequence isn't decreasing for all $n > 3$. The contradiction tells you the sequence decreases for some $n > 3$. That's not the same as saying the sequence decreases for all $n > 3$.

@Hall you did not respond to this...i would be interested on your view on this...

 

[nuuskur]

@fresh_42 ...you had mentioned in my other post that I cannot use L' Hopital's rule...kindly clarify...cheers.

The rule is applicable for differentiable functions. You can conclude with L'Hopital that
$$\lim _{x\to\infty} \frac{\ln x^2}{x} = 0.$$
In the discrete case this is also true due to continuity.

You should be careful, though. You claimed something like
$$\lim _{x\to a} f(x) = L \Rightarrow (a_n \to a \Rightarrow \lim _n f(a_n) = L).$$
This is true for continuous functions. To be more precise, the function of interest has to be continuous around the point of convergence. In the event $x\to\infty$, it should be that $f$ is continuous from some point onward.

 

[chwala]

The rule is applicable for differentiable functions. You can conclude with L'Hopital that
$$\lim _{x\to\infty} \frac{\ln x^2}{x} = 0.$$
In the discrete case this is also true due to continuity.

You should be careful, though. You claimed something like
$$\lim _{x\to a} f(x) = L \Rightarrow (a_n \to a \Rightarrow \lim _n f(a_n) = L).$$
This is true for continuous functions. To be more precise, the function of interest has to be continuous around the point of convergence. In the event $x\to\infty$, it should be that $f$ is continuous from some point onward.

You state that the rule applies to differentiable functions? This function is not continous at $x=0$, do we still consider it as a differentiable function?

In my understanding, if a function is not continous then it is not differentiable.

 

[pasmith]

You state that the rule applies to differentiable functions? This function is not continous at $x=0$, do we still consider it as a differentiable function?

In my understanding, if a function is not continous then it is not differentiable.

The limit is being taken as $x \to \infty$; the behaviour of the function at $x = 0$ is of no consequence.