This looks like an alternative series because cosine value continues to move negative and positive, but it also makes sense to me that this isn't an alternative test. In this case, I'm lost.
#Which method do I use to find convergence?
proud cape
lethal basinBOT
This looks like an alternative series because cosine value continues to move neg...
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wispy flame
This looks like an alternative series because cosine value continues to move neg...
hint : it converges absolutely, prove how
proud cape
According to Absolutely Convergent test, it says that if I put in absolute value between the series
and if I get that the answer is convergent, then the series is absolutely convergent
That's what I know
but application is hard here
because i don't think cos(n) can be exterminated from the equation here
wispy flame
because i don't think cos(n) can be exterminated from the equation here
|cos(n)|<=1
proud cape
how does that work?
are you putting in 1 inside n?
0.717?
or are we just ignoring cos(n) here?
wispy flame
or are we just ignoring cos(n) here?
|x cos(n)|<=|x|
proud cape
I'm really sorry, but I don't understand it. Could you elaborate that for me?
raw swallow
or are we just ignoring cos(n) here?
We are not ignoring cos(n), we are considering the extrema of cos(x)
which are bounds as well
grim flower
then we left with 1/n^2?
Yeah, which is larger than |cos n|/n²
So what can you tell about the series of |cos n|/n² ?
Given that it is well known that the series of 1/n² converges
proud cape
it is also convergent as well
but still can't go over how |cos n|/n^2 is gone for now
grim flower
Well the reasoning is as follows
- It suffices to show that the series of |cos n|/n² converges to show that the series of cos n /n² converges
This is called absolute convergence
- You are doing exactly that now, by comparison test
grim flower
but still can't go over how |cos n|/n^2 is gone for now
You get the gist?
proud cape
I understand that we are using cos(n)/n^2 to prove that this is convergent
But how do we get to 1/n^2? Do we take out cos(n) because it making the answer switching from positive to negative in the sequnce?
grim flower
No, we have no idea how to compute the series to show it converges
So instead we do a comparison test
If a series of positive terms is smaller than another one that converges, then it converges
And here the bigger one that converges is the series of 1/n²
grim flower
so we're using Diret Comparison Test?
To show that the series of |cos n|/n² converges
Because the terms are all positive
And then, to show that the series of cos n/n² converges, you use the knowledge that the series of |cos n|/n² converges
proud cape
ok so i get that |cos n|/n^2 converges and that directly makes cos n/n^2 convergent
But how do we know that |cos n|/n^2 is convergent?
It's a calculation that is stopping me
grim flower
As said, comparison test
proud cape
Limit Comparison Test?
grim flower
You can't compute that series
So you will take a bigger series
And show it converges
proud cape
Ok i get it now
so we can't directly compute series with cos(n)
so we compute 1/n^2
which we get that it's convergent
proud cape
Wait hold on give me a moment
According to the definition of Limit Comparison Test
Which states that an > 0 and bn > 0
for all n
if lim n->(infinity) an/bn = L.
grim flower
It's not limit comparison, it's just comparison
grim flower
No
Ok, think of it like this
Take A_n the sum of a_i from 1 to n
You know that A_n is nondecreasing
proud cape
A_n here is cos(n) here right?
actually
cos(n)/n^2
just the original series equation
proud cape
ok
grim flower
And it is upper bounded by B_n = sum of b_i for from 1 to n
B_n is nondecreasing and known to converge at a limit B
So A_n <= B_n <= B
So since A_n is upper bounded and nondecreasing
It converges
That is just a plain comparison test that you can perform when you know that 0 <= a_i <= b_i and B_n converges
grim flower
Now we just need to pick the right a_n and b_n
b_n is the larger one which series is known to converge
So 1/n²
proud cape
ok
grim flower
So which one is the a_n?
proud cape
cos(n)/n^2
proud cape
but can we put absolute value just because we need it to be positive?
grim flower
Well your subgoal is to show that the series of |cos n|/n² converges
Because it is sufficient
proud cape
what is sufficient?
grim flower
It is sufficient to know that the series of |cos n|/n² converges
To show that the series of cos n / n² also converges
proud cape
ok
raw swallow
Do you need extra justification on Rion's last statement?
proud cape
actually yes because I'm still not sure why do we suddenly put absolute value sign on cos(n) because the original question is cos(n) (i know that to do direct comparison test we need to make it above 0)
grim flower
I would also appreciate if Rafain can take over on that, I am in a train and the wifi is not all that good right now
proud cape
thank you for the effort so far, sorry for not being able to understand the explanation
grim flower
Do you need extra justification on Rion's last statement?
Can I request you to do that?
raw swallow
Can I request you to do that?
My pleasure
Note that |cos(n)|/n^2 = |cos(n)/n^2|
Hence the convergence of the series sum{|cos(n)|/n^2} is, indeed, the absolute convergence of the series sum{cos(n)/n^2}
proud cape
I see that we need to put in the absolute value sign in order to prove that the series is absolutely convergent or divergent
raw swallow
I see that we need to put in the absolute value sign in order to prove that the ...
The end goal is that, yes. More aptly put, the absolute value sign was necessary for applying Direct Comparison test, so as to prove that sum{|cos(n)|/n^2} is convergent.
proud cape
Why is it necessary? Is it because cos(n) sometimes give negative value?
proud cape
Ok so first we apply the direct comparison rule with absolute value and the find that this is convergent.
Wait what's the difference between the Comparison Test and Absolute Convergent Test?
Only difference I see is that Absolute Convergent test is we use absolute value
raw swallow
The second half is an application of the Absolute Convergence test.
It states that if sum{|a_n|} converges, then sum{a_n} converges too.
You can establish this e.g. via Cauchy criterion
proud cape
I may have wrote the wrong definition for Absolute Convergent Test
On my note is says that {a_n} is absolutely convergent is the series of absolute value is convergent
Doesn't talk about comparison at all
raw swallow
That is just the definition of absolute convergence
proud cape
oh ok
The guy on my explanation video (taking calc class online) came with example question of (-1)^n+1/n^2
And then he put absolute value in the series and then eliminate (-1)^n+1 since all values will be positive and we don't need alternating value in this case
He gets 1/n^2 and it converges
Hence the series converges
raw swallow
Yes, that is an application of the Absolute Convergence test
proud cape
Are we not using Direct Comparison because of cos(n)?
Becuase cos(n) changes its signs right?
as n increases?
raw swallow
We are using both tests
raw swallow
1/n^2
-> |cos(n)|/n^2 by DCT
-> cos(n)/n^2 by ACT
proud cape
ok
grim flower
The reason why we are using the direct comparison test is because you want to show that the sum of |cos n|/n² converges (this is your prerequisite for the absolute convergence test), but you do not know if it does from the get-go
And it is not a sum that can be explicitly computed for you, so you will deal with it by direct comparison
proud cape
ok
grim flower
In your video, no comparison test was necessary because the author already had the knowledge that the series of 1/n² is convergent
proud cape
so since we don't know if cos(n)/n^2 is convergent or not, we're applying the ACT before DCT?
grim flower
You anticipate using the ACT
You will use the DCT first with |cos n|/n²
And then, you apply the ACT with cos n / n²
proud cape
ok
grim flower
I hope that it is clearer for you?