Determine Convergence/Divergence of Sequence: f(x)=ln(x)^2/x

[chwala]

Homework Statement

Determine the convergence or divergence of the sequence $a_n= \left[\dfrac {\ln (n)^2}{n}\right]$

Relevant Equations

convergence knowledge

$a_n= \left[\dfrac {\ln (n)^2}{n}\right]$

We may consider a function of a real variable. This is my approach;

$f(x) =\left[\dfrac {\ln (x)^2}{x}\right]$

Applying L'Hopital's rule we shall have;

$\displaystyle\lim_ {x\to\infty} \left[\dfrac {\ln (x)^2}{x}\right]=\lim_ {x\to\infty}\left[ \dfrac {2}{x}\right]=0$

because $f(n)=a_n$ for every positive integer $n$, then we may conclude that the sequence converges to $0$.

I would appreciate any insight on this...cheers.

 

[Mark44]

Homework Statement:: Determine the convergence or divergence of the sequence $a_n= \left[\dfrac {\ln (n)^2}{n}\right]$
Relevant Equations:: convergence knowledge

$a_n= \left[\dfrac {\ln (n)^2}{n}\right]$

We may consider a function of a real variable. This is my approach;

$f(x) =\left[\dfrac {\ln (x)^2}{x}\right]$

Applying L'Hopital's rule we shall have;

$\lim {x→∞} \left[\dfrac {\ln (x)^2}{x}\right]=\lim {x→∞}\left[ \dfrac {2}{x}\right]=0$

because $f(n)=a_n$ for every positive integer $n$, then we may conclude that the sequence converges to $0$.

I would appreciate any insight on this...cheers.

Looks fine to me.

 

[chwala]

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

 

[fresh_42]

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

You can, just explain how. Wikipedia's version says
$$
\lim_{x \to x_0}\dfrac{f'(x)}{g'(x)}=c \Longrightarrow \lim_{x \to x_0}\dfrac{f(x)}{g(x)}=c
$$
This seems to be different from what you have used.

 

[WWGD]

I'd suggest the use of parentheses to avoid ambiguity. $ln(x)^2=ln(x^2)$ or $(ln(x))^2$?

 

[WWGD]

But notice $ln(e^x)=x$. And see the behavior of $\frac {x}{e^x}$. Or even $\frac {x^2}{e^x}$.

 

[Mark44]

I assumed in my reply that this:

$a_n= \left[\dfrac {\ln (n)^2}{n}\right]$

means this:
$a_n= \left[\dfrac {(\ln (n))^2}{n}\right]$

 

[WWGD]

I assumed in my reply that this:

means this:
$a_n= \left[\dfrac {(\ln (n))^2}{n}\right]$

I understand , but it seems it could be interpreted either way without further explanation

 

[SammyS]

$f(x) =\left[\dfrac {\ln (x)^2}{x}\right]$

Applying L'Hopital's rule we shall have;

$\lim {x→∞} \left[\dfrac {\ln (x)^2}{x}\right]=\lim {x→∞}\left[ \dfrac {2}{x}\right]=0$

Assuming what @Mark44 assumed above:

$f(x) =\dfrac { \left(\ln(x)\right)^2}{x}$

You need to apply L'Hopital's Rule twice.

$\displaystyle \lim _{x\to\infty} \dfrac {\ln (x)^2}{x}= \lim _{x\to\infty} \dfrac {2\ln (x)}{x}= \lim _{x\to\infty} \dfrac {2}{x}=0$

 

[Mark44]

I understand , but it seems it could be interpreted either way without further explanation

If it could be interpreted in more than one way, explanation is needed.

 

[chwala]

I'd suggest the use of parentheses to avoid ambiguity. $ln(x)^2=ln(x^2)$ or $(ln(x))^2$?

I posted exactly as it appears on textbook...

 

[fresh_42]

I posted exactly as it appears on textbook...

I don't think your notation is ambiguous. Actually, $\log^2(n)$ which is often used instead is ambiguous. To me this would clearly be $\log^2(n)=\log(\log(n))$ but people use it for $\log^2(n)=(\log(n))^2.$ As there is no chance to read your notation as $\log(n^2),$ it is obviously the square of $\log(n).$

How you applied L'Hôpital and why you can switch from discrete to continuous are the mathematical problems here. No big deal, but it deserves an explanation. Not the notation.

 

[WWGD]

I posted exactly as it appears on textbook...

Don't mean to blame you, just to add that in some cases it may be misunderstood.

 

[SammyS]

Limit of a Sequence
Let $f$ be a function of a real variable such that

$\displaystyle\lim_{x\to\infty} f(x)= L.$

If ${a_n}$ is a sequence such that $f(n) = a_n$ for every positive integer $n$, then

$\lim_{n\to\infty} {a_n} = L.$

unless you want to dispute the theorem...my application of L'Hopital's rule is straightforward.

I corrected some typos in the quoted text above.

I agree with @fresh_42 . You post these problems with your solutions, many times as a review. You apparently are asking for comment.
My general comment is: For you to get the most out of your review, you should include much more in regards to explanation of what you are doing.

In this case, it isn't at all clear that you applied L'Hopital properly.

For one thing, what allows you to apply L'Hopital's rule?

For another, what is your result for $\displaystyle \frac{d}{dx} \ln(x)^2 \ ?$

 

[Mark44]

To me this would clearly be $\log^2(n)=\log(\log(n))$ but people use it for $\log^2(n)=(\log(n))^2.$

Using the same logic, would you interpret $\sin^2(x)$ to mean $\sin(\sin(x))$?

 

[chwala]

I corrected some typos in the quoted text above.

I agree with @fresh_42 . You post these problems with your solutions, many times as a review. You apparently are asking for comment.
My general comment is: For you to get the most out of your review, you should include much more in regards to explanation of what you are doing.

In this case, it isn't at all clear that you applied L'Hopital properly.

For one thing, what allows you to apply L'Hopital's rule?

For another, what is your result for $\displaystyle \frac{d}{dx} \ln(x)^2 \ ?$

@SammyS hello...i did not have a solution for this particular problem ...otherwise i would have mentioned that (as i always do).

Nevertheless, the L'Hopital's rule will involve use of chain rule on the numerator...

i.e let $f(x)= \ln (x^2)$ and $g(x) = x$ reference L'Hopital rule post $4$.

$f^{'}(x)=\left[ \dfrac{1}{x^2} ⋅2x\right]= \left[\dfrac{2}{x}\right]$ and $g^{'}( x)=1$

...the steps to solution will follow thereafter after taking limits...

Cheers and thanks all for your input.

 

[chwala]

I corrected some typos in the quoted text above.

I agree with @fresh_42 . You post these problems with your solutions, many times as a review. You apparently are asking for comment.
My general comment is: For you to get the most out of your review, you should include much more in regards to explanation of what you are doing.

In this case, it isn't at all clear that you applied L'Hopital properly.

For one thing, what allows you to apply L'Hopital's rule?

For another, what is your result for $\displaystyle \frac{d}{dx} \ln(x)^2 \ ?$

It would be helpful if there is another approach to this problem.

 

[fresh_42]

Using the same logic, would you interpret $\sin^2(x)$ to mean $\sin(\sin(x))$?

I get used to the convention $\sin^2(x)=(\sin(x))^2$ and I use it since it is convenient. The multiplication of functions as consecutive applications is so rare in calculus that it needs an extra definition anyway when it is used. Even in CS, people write $\log\log$ and not $\log^2.$

The current convention reduces the number of symbols that have to be used $(\log(x))^2=\log^2x$ and $\log(x^2)=\log x^2$ which is a good justification. I even would write $\log\log x$ for its clearance rather than e.g. $\log^{(2)}x$ which could be read as $\dfrac{d^2}{dx^2}\log(x).$

My comment was a polemic against the discussion of notation here. The notation here was $\log(n)$ and the parentheses resolve any possible ambiguity in my opinion that occurs in $\log n$ where you cannot simply add a square at the end without changing the meaning. I quoted the L'Hôpital rule from Wikipedia (here or in the previous thread) and asked how he applied it since I do not see it. I asked him therefore which version he used. I think that deserves an answer.

... my application of L'Hopital's rule is straightforward.

is none and I will refrain from further participation in @chwala's threads from now on besides moderation issues anyway, so this question is closed from my point of view.

 

[chwala]

I get used to the convention $\sin^2(x)=(\sin(x))^2$ and I use it since it is convenient. The multiplication of functions as consecutive applications is so rare in calculus that it needs an extra definition anyway when it is used. Even in CS, people write $\log\log$ and not $\log^2.$

The current convention reduces the number of symbols that have to be used $(\log(x))^2=\log^2x$ and $\log(x^2)=\log x^2$ which is a good justification. I even would write $\log\log x$ for its clearance rather than e.g. $\log^{(2)}x$ which could be read as $\dfrac{d^2}{dx^2}\log(x).$

My comment was a polemic against the discussion of notation here. The notation here was $\log(n)$ and the parentheses resolve any possible ambiguity in my opinion that occurs in $\log n$ where you cannot simply add a square at the end without changing the meaning. I quoted the L'Hôpital rule from Wikipedia (here or in the previous thread) and asked how he applied it since I do not see it. I asked him therefore which version he used. I think that deserves an answer.

is none and I will refrain from further participation in @chwala's threads from now on besides moderation issues anyway, so this question is closed from my point of view.

@fresh_42

I hope I haven't done anything to agitate you...if so kindly accept my apologies sir...you are a great mentor and I always look forward to your insight.

Regards,
Chwala

 

[Mark44]

I quoted the L'Hôpital rule from Wikipedia (here or in the previous thread) and asked how he applied it since I do not see it.

I agree that it wasn't immediately clear -- I had to work through the two applications of L'Hopital to arrive at the result @chwala showed.

 

[chwala]

I agree that it wasn't immediately clear -- I had to work through the two applications of L'Hopital to arrive at the result @chwala showed.

All my mistake...apologies, learning point for me...I will endeavour to post all necessary steps in my future posts.

 

[Mark44]

All my mistake...apologies, learning point for me...I will endeavour to post all necessary steps in my future posts.

Or if you take shortcuts, at least describe what you did.

 

[vela]

I assumed in my reply that this:

means this:
$a_n= \left[\dfrac {(\ln (n))^2}{n}\right]$

That's funny. When I saw the problem in the other thread, I interpreted it the other way. The fact that the parentheses don't include the log, to me, means you should square what's inside the parentheses and then take the log.

Coincidentally, it turns out it didn't matter because
$$\lim_{x\to\infty} \frac{\log (x^2)}{x} = \lim_{x\to\infty} \frac{2 \log x}{x} = \lim_{x\to\infty} \frac{2/x}{1} =\lim_{x\to\infty} \frac{2}{x}.$$ Either way, you end up with that last limit.

 

[chwala]

My last question on this...does it therefore mean that we can always treat sequences as functions when determining convergence or divergence? Thanks...

 

[WWGD]

My last question on this...does it therefore mean that we can always treat sequences as functions when determining convergence or divergence? Thanks...

In what way? A function by itself doesn't converge or diverge. A sequence of function may converge or diverge. Though strictly speaking, a sequence is a function whose domain is the Natural Numbers.

 

[chwala]

In what way? A function by itself doesn't converge or diverge. A sequence of function may converge or diverge. Though strictly speaking, a sequence is a function whose domain is the Natural Numbers.

Yes, my question was specifically on whether we can treat sequences as functions...ofcourse the domain would be the set of Natural numbers.

 

[Hall]

@chwala We can prove, without resorting to derivatives and L'Hopitals rule, that $\lim a_n =0$.
Let us write down the first few terms of the sequence
$$
\begin{align*}
a_1 = 0 \\
a_2 = 0.69314\\
a_3 = 0.7324\\
a_4 = 0.69314\\
a_5 = 0.64377\\
a_6 = 0.59725\\
\end{align*}
$$
It seems that the sequence is decreasing after $n=3$. Let's prove that it is indeed the case.

Assume that $a_n = \frac{2 ln(n)}{n}$ is not decreasing after $n=3$, so, we have
$$
\begin{align*}
\frac{ln (n)}{n} \leq \frac{ ln (n+1)}{n+1} \\
ln ( n^{n+1}) \leq ln ((n+1)^{n})\\
n^{n+1} \leq (n+1)^{n}\\
\textrm{ take n=5} \\
15,625 \leq 7,776\\
\textrm{the above is not true, hence our assumption was wrong}\\
\textrm{Thus, our sequence is decreasing after n=3}\\
\end{align*}
$$
Now, we have for monotone decreasing sequence
$$
\lim a_n = \inf\{ a_n\}
$$
Thus, $\lim a_n = 0$

 

[WWGD]

@chwala We can prove, without resorting to derivatives and L'Hopitals rule, that $\lim a_n =0$.
Let us write down the first few terms of the sequence
$$
\begin{align*}
a_1 = 0 \\
a_2 = 0.69314\\
a_3 = 0.7324\\
a_4 = 0.69314\\
a_5 = 0.64377\\
a_6 = 0.59725\\
\end{align*}
$$
It seems that the sequence is decreasing after $n=3$. Let's prove that it is indeed the case.

Assume that $a_n = \frac{2 ln(n)}{n}$ is not decreasing after $n=3$, so, we have
$$
\begin{align*}
\frac{ln (n)}{n} \leq \frac{ ln (n+1)}{n+1} \\
ln ( n^{n+1}) \leq ln ((n+1)^{n})\\
n^{n+1} \leq (n+1)^{n}\\
\textrm{ take n=5} \\
15,625 \leq 7,776\\
\textrm{the above is not true, hence our assumption was wrong}\\
\textrm{Thus, our sequence is decreasing after n=3}\\
\end{align*}
$$
Now, we have for monotone decreasing sequence
$$
\lim a_n = \inf\{ a_n\}
$$
Thus, $\lim a_n = 0$

How do you prove 0 is the inf? Maybe the sequence decreases increasingly slowly.

 

[Hall]

How do you prove 0 is the inf? Maybe the sequence decreases increasingly slowly.

$0 \leq \frac{2 ln(n)}{n} \leq \frac{2 ln(3)}{3}$

 

[vela]

Assume that $a_n = \frac{2 ln(n)}{n}$ is not decreasing after $n=3$, so, we have
$$
\begin{align*}
\frac{ln (n)}{n} \leq \frac{ ln (n+1)}{n+1} \\
ln ( n^{n+1}) \leq ln ((n+1)^{n})\\
n^{n+1} \leq (n+1)^{n}\\
\textrm{ take n=5} \\
15,625 \leq 7,776\\
\textrm{the above is not true, hence our assumption was wrong}\\
\textrm{Thus, our sequence is decreasing after n=3}\\
\end{align*}
$$

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]

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.