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

• chwala
In summary: L'Hopital's rule is just an alternative method of solving the problem...it should not be an issue if not applied.But, I really appreciate the input.@fresh_42 ...you had mentioned in my other post that I cannot use L' Hopital's rule...kindly clarify...cheers.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'H
chwala
Gold Member
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.

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vanhees71
chwala said:
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.

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

chwala said:
@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.

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

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

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]##

Mark44 said:
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

vela
chwala said:
##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##

vanhees71 and Mark44
WWGD said:
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.

WWGD
WWGD said:
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...

chwala said:
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.

vanhees71 and SammyS
chwala said:
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.

Hall, chwala and fresh_42
chwala said:
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 \ ?##

fresh_42 said:
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))##?

SammyS said:
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.

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SammyS said:
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.

Mark44 said:
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.
chwala said:
... 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.

fresh_42 said:
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

fresh_42 said:
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.

Mark44 said:
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.

chwala said:
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.

chwala
Mark44 said:
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.

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

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chwala said:
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.

Hall and chwala
WWGD said:
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.

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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##

Hall said:
@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.

WWGD said:
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}##

Hall said:
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##.

chwala
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*}

vela said:
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...

chwala said:
@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.

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chwala
nuuskur said:
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.

chwala said:
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.

chwala and nuuskur

## What is the definition of convergence and divergence in a sequence?

Convergence in a sequence means that the terms of the sequence approach a fixed value as the number of terms increases. Divergence, on the other hand, means that the terms of the sequence do not approach a fixed value and instead either increase or decrease without bound.

## What is the formula for determining the convergence or divergence of a sequence?

The formula for determining the convergence or divergence of a sequence is lim n→∞ an = L, where L is the limit of the sequence as n approaches infinity. If the limit exists and is a finite number, the sequence is convergent. If the limit does not exist or is infinite, the sequence is divergent.

## How do you determine the convergence or divergence of a function using the ratio test?

The ratio test is a method for determining the convergence or divergence of a sequence. To use this test, you take the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term. If this limit is less than 1, the sequence is convergent. If the limit is greater than 1, the sequence is divergent. If the limit is equal to 1, the test is inconclusive and another method must be used.

## Can the convergence or divergence of a sequence be affected by the choice of x in the function f(x)=ln(x)^2/x?

Yes, the convergence or divergence of a sequence can be affected by the choice of x in the function f(x)=ln(x)^2/x. For example, if x=1, the sequence will be convergent, but if x=0, the sequence will be divergent. It is important to consider the domain of the function when determining the convergence or divergence of a sequence.

## What are some other methods for determining the convergence or divergence of a sequence?

Other methods for determining the convergence or divergence of a sequence include the integral test, comparison test, and root test. These tests compare the given sequence to a known convergent or divergent sequence to determine its behavior. Additionally, the alternating series test can be used for alternating sequences and the direct comparison test can be used for non-negative sequences.

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