Takeover Season
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If we look at the following statement:

"Also if |r| > 1, then the sequence (r^n) n∈N has the subsequence (r^{2n}) n∈N which is not bounded, and hence not convergent. Consequently (r^n) n∈N diverges, and hence the series diverges."

Can someone just confirm if my intuitions are correct and answer some questions that I have please:

1. Since |r| > 1, surely as n --> ∞, the sequence (r^n)'s n∈N terms get larger and larger and are unbounded. So, (r^n) n∈N is an unbounded sequence, why did it talk about the subsequence (r^{2n}) n∈N instead?

2. Now, regarding the actual ideas. If someone could just confirm please:

If a sequence is convergent, then it must be bounded. Using the contrapositive: If a sequence is not bounded, it is divergent. So, as a subsequence is a sequence itself, the subsequence is unbounded and therefore not convergent.

So, in our case, as the subsequence is not bounded, it is not convergent.

If a sequence of real numbers converges, then all of its subsequences converge too (and to the same limit of the original sequence). Using the contrapositive: If there are any two subsequences that converge to different limits OR if there is any subsequence that diverges, then the sequence of real numbers diverges.

So, in our case, the subsequence diverges, hence the sequence diverges.

3. Now, \sum _{n=1}^{\infty } r^n is the series where r^n are the terms/entries of the sequence (r^n) n∈N. If the sequence diverges, then why does this mean the series diverges?

Thank you
Last edited by Takeover Season; 10 months ago
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ghostwalker
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(Original post by Takeover Season)
If we look at the following statement:

"Also if |r| > 1, then the sequence (r^n) n∈N has the subsequence (r^{2n}) n∈N which is not bounded, and hence not convergent. Consequently (r^n) n∈N diverges, and hence the series diverges."

Can someone just confirm if my intuitions are correct and answer some questions that I have please:

1. Since |r| > 1, surely as n --> ∞, the sequence (r^n)'s n∈N terms get larger and larger and are unbounded. So, (r^n) n∈N is an unbounded sequence, why did it talk about the subsequence (r^{2n}) n∈N instead?
There's no real requirement to use the subsequence - perhaps it's just included to show how it might be used.

(r^{2n})_{n\in\mathbb{N}} would be an increasing sequence of positive terms which is unbounded above, but bounded below by 0.

(r^n)_{n\in\mathbb{N}} is unbounded above AND may also be unbounded below if r<-1


If a sequence is convergent, then it must be bounded. Using the contrapositive: If a sequence is not bounded, it is divergent. So, as a subsequence is a sequence itself, the subsequence is unbounded and therefore not convergent.

So, in our case, as the subsequence is not bounded, it is not convergent.
Yes.

If a sequence of real numbers converges, then all of its subsequences converge too (and to the same limit of the original sequence). Using the contrapositive: If there are any two subsequences that converge to different limits OR if there is any subsequence that diverges, then the sequence of real numbers diverges.

So, in our case, the subsequence diverges, hence the sequence diverges.
OK

3. Now, \sum _{n=1}^{\infty } r^n is the series where r^n are the terms/entries of the sequence (r^n) n∈N. If the sequence diverges, then why does this mean the series diverges?
Well, what's the requirement for a series to converge? What criterion must the terms of the sequence satisfy?


By the way, are you working through a book, or is this from lectures, or something else. As, you're asking a lot of basic questions that any introductory text should cover.
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Takeover Season
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(Original post by ghostwalker)
There's no real requirement to use the subsequence - perhaps it's just included to show how it might be used.

(r^{2n})_{n\in\mathbb{N}} would be an increasing sequence of positive terms which is unbounded above, but bounded below by 0.

(r^n)_{n\in\mathbb{N}} is unbounded above AND may also be unbounded below if r<-1




Yes.



OK



Well, what's the requirement for a series to converge? What criterion must the terms of the sequence satisfy?


By the way, are you working through a book, or is this from lectures, or something else. As, you're asking a lot of basic questions that any introductory text should cover.
Thank you for your reply, when you say "OK", can I take that as meaning it is correct?

For a series which is a sum of the terms of the sequence (a_n), for this series to converge, it must be that the partial sum of the sequence (a_n) converges.

Oh right, so it is a definition and we can say:
If the series converges, then the partial sum of the sequence converges. So, the contrapositive says, if the partial sum of the sequence diverges, then the series diverges?

Sometimes it gives equivalent statements, like if the series converges, then its value is the limit as n--> infinity of the partial sum, so the partial sum must converge for the series to converge. Can I take that as a definition and so write it as an if-then / if and only if statement?

I am working through lectures, but I am struggling on this course as I wasn't good on the first year course version.
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ghostwalker
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(Original post by Takeover Season)
Thank you for your reply, when you say "OK", can I take that as meaning it is correct?
Yes

For a series which is a sum of the terms of the sequence (a_n), for this series to converge, it must be that the partial sum of the sequence (a_n) converges.
The partial sums of the series must converge.

I assume you mean:

Series is a1+a2+a3+...+an+... formed from the sequence a1, a2, a3, ..., an, ...
Partial sums form a sequence in their own right:
s1, s2, s3, ...

where
s1=a1
s2+a1+a2
s3=a1+a2+a3

Oh right, so it is a definition and we can say:
If the series converges, then the partial sum of the sequence converges. So, the contrapositive says, if the partial sum of the sequence diverges, then the series diverges?
Replace sequence with series.

Sometimes it gives equivalent statements, like if the series converges, then its value is the limit as n--> infinity of the partial sum, so the partial sum must converge for the series to converge. Can I take that as a definition and so write it as an if-then / if and only if statement?
No, it's not a definition, it's simply restating what it means for a series to converge:

A series converges (as a series), means the partial sums converge as a sequence.

I would strongly recommend going back and getting a firm handle on sequences and series. Your posts are very difficult to follow.
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ghostwalker
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Addendum:

I could be misinterpreting some of what you're saying - it's difficult to know at times.


(can't edit my previous post due to a bug in TSR relating to quotes being split up)
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(Original post by ghostwalker)
Yes



The partial sums of the series must converge.

I assume you mean:

Series is a1+a2+a3+...+an+... formed from the sequence a1, a2, a3, ..., an, ...
Partial sums form a sequence in their own right:
s1, s2, s3, ...

where
s1=a1
s2+a1+a2
s3=a1+a2+a3



Replace sequence with series.



No, it's not a definition, it's simply restating what it means for a series to converge:

A series converges (as a series), means the partial sums converge as a sequence.

I would strongly recommend going back and getting a firm handle on sequences and series. Your posts are very difficult to follow.
Thank you, you have clarified everything. I am sorry, I am really stupid . I am going back to the chapter to revise it now.
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ghostwalker
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(Original post by Takeover Season)
Thank you, you have clarified everything. I am sorry, I am really stupid . I am going back to the chapter to revise it now.
Regardless of ability, analysis is quite a difficult subject when you first encounter it (totally unlike school mathematics), and it pays dividends to take the time to fully understand and appreciate the definitions and styles of proof. Don't beat yourself up over it.
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Takeover Season
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(Original post by ghostwalker)
Regardless of ability, analysis is quite a difficult subject when you first encounter it (totally unlike school mathematics), and it pays dividends to take the time to fully understand and appreciate the definitions and styles of proof. Don't beat yourself up over it.
Thank you for the support and kind words
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(Original post by Takeover Season)
Thank you, you have clarified everything. I am sorry, I am really stupid . I am going back to the chapter to revise it now.
You're not stupid, lots of people (including myself) found the jump up to undergraduate mathematics difficult but most adjust alright in the end. It must be especially hard if you are learning by yourself without peers or lecturers/supervisors to discuss it with.
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(Original post by 15Characters...)
You're not stupid, lots of people (including myself) found the jump up to undergraduate mathematics difficult but most adjust alright in the end. It must be especially hard if you are learning by yourself without peers or lecturers/supervisors to discuss it with.
Thank you so much for the support and kind words . Yes, it is difficult in these times with online learning.
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