How do I show the following sequence is bounded above ?

Sophdof1

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I want to show the above ^

Any ideas?
 

Dr.Peterson

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Each term is itself a sum of n terms. What is the largest of those terms? What does that tell you about an upper bound on the sum?
 

Sophdof1

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Each term is itself a sum of n terms. What is the largest of those terms? What does that tell you about an upper bound on the sum?
It’s an infinite number of sums
 

Dr.Peterson

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Don't worry about all of them at once; focus on just one term, \(\displaystyle a_n\). That is a finite sum. You just have to find a number that that one sum is less than. (It will turn out to be a number that they are all less than!) Give it a try. Follow my suggestion.
 

Sophdof1

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Don't worry about all of them at once; focus on just one term, \(\displaystyle a_n\). That is a finite sum. You just have to find a number that that one sum is less than. (It will turn out to be a number that they are all less than!) Give it a try. Follow my suggestion.

Should I just take random numbers and trial them?
 

pka

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I want to show the above ^ Any ideas?
Do you know about the harmonic numbers \(\displaystyle H_n~?\) and the the digamma function. Here are two webpages One & Two.
The nth harmonic number is \(\displaystyle {H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} \) and your sequence turns out to be \(\displaystyle a_n=H_{2n}-H_n\)
On the two references, you can find links to other research do by John Conway and others.
 

Dr.Peterson

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No, just answer my questions! What is the largest term in the sum?

It may help (or may not) if you first consider a particular value of n, just to get a more human image of what this is. \(\displaystyle a_4\), for example, is 1/5 + 1/6 + 1/7 + 1/8. What is the largest term? If you replaced each term by that one, would the resulting sum be larger? Do the same thing to the general term, \(\displaystyle a_n\).

One way or another, I need to see you do some work!
 

Sophdof1

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I
No, just answer my questions! What is the largest term in the sum?

It may help (or may not) if you first consider a particular value of n, just to get a more human image of what this is. \(\displaystyle a_4\), for example, is 1/5 + 1/6 + 1/7 + 1/8. What is the largest term? If you replaced each term by that one, would the resulting sum be larger? Do the same thing to the general term, \(\displaystyle a_n\).

One way or another, I need to see you do some work!
I understand that logically - but I don’t understand how I can actually go about proving this via implications. Everything you said makes sense but I am wanting to prove this thoroughly and my proof just looks messy and not thorough enough.
 

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Dr.Peterson

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You did what I had in mind, but didn't go far enough. You want to show that \(\displaystyle a_n\) is always less than some fixed number. You've only shown, so far, that it is less than a function of n. What fixed number is that function always less than?
 

Sophdof1

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Is there any way you can maybe lead me further to getting this number? I don’t have a clue
 

Jomo

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Your work is perfect but a little incomplete. n/(n+1) < what number for any positive number n
If you really do not see this, then plug in positive numbers for n and see what number than are all less than.
 

Sophdof1

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Your work is perfect but a little incomplete. n/(n+1) < what number for any positive number n
If you really do not see this, then plug in positive numbers for n and see what number than are all less than.
I can see it is 1. But I don’t know how one can actually prove that it is 1...
 

Dr.Peterson

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Surely you can show that n/(n+1) < 1 for all n. But you don't really need to do that.

Rather than use the fact that each term is no more than 1/(n+1), just use the fact that each term is less than 1/n!
 

Jomo

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I can see it is 1. But I don’t know how one can actually prove that it is 1...
No, I disagree with you saying that you can see that n/(n+1) < 1. It sounds to me that you think that the only number that n/(n+1) is less than is 1. n/(n+1) is also less then 3/2, sqrt(7), pi and the number e.
 

Jomo

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I can see it is 1. But I don’t know how one can actually prove that it is 1...
n/(n+1) ? 1 for positive n
n ? (n+1) for positive n
Now this last line is n <(n+1)
So the 1st line is n/(n+1) < 1
OK??
 

Sophdof1

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Surely you can show that n/(n+1) < 1 for all n. But you don't really need to do that.

Rather than use the fact that each term is no more than 1/(n+1), just use the fact that each term is less than 1/n!
You seem to have confused me now. Is there any way you can explain this in a different angle?
 

Jomo

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Same angle: an= 1/(n+1) + 1/(n+2) + 1/(n+3) + 1/(n+4) < 1/n + 1/n 1/n + 1/n =4/n
Continue....


Well actually an= 1/(n+1) + 1/(n+2) + 1/(n+3) + 1/(n+4) + .... + 1/2n
So an= 1/(n+1) + 1/(n+2) + 1/(n+3) + 1/(n+4) + .... + 1/2n < 1/(n+1) 1/(n+1) + 1/(n+1) + ... + 1/(n+1) =n/(n+1)
 
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Dr.Peterson

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You seem to have confused me now. Is there any way you can explain this in a different angle?
There are two angles I've suggested: You can just continue what you did to get to the final answer, for which we've given you plenty of guidance; or you can back up and redo it, replacing 1/(n+1) with 1/n in your n terms. [I see Jomo just wrote about that, but got it wrong - don't let that confuse you.] You don't need to attempt the latter if you don't see my point. Just finish the work you did, and show us your conclusion.

Then we can see whether there is anything lacking in what you write. I think it will be a perfectly good proof.

Jomo showed you one way to finish, using the fact that n < n+1 for all n; another way would be to say that n/(n+1) = [(n+1) - 1]/(n+1) = 1 - 1/(n+1) < 1 for all non-negative n.

One more point that Jomo hinted at: 1 is not the only number that will work; it may not even be the lowest possible upper bound. All you need is some upper bound. So don't expect to find the answer. But our answer of 1 is a very nice upper bound.
 

Sophdof1

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Okay so I got my proof!

I’m pretty certain I have done it now.
 

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Dr.Peterson

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Looks like you've got it.
 
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