Is my proof for Absolute convergence => Convergence right?

  • #1
Is my proof for "Absolute convergence => Convergence" right?

I had to prove it for my calc test today; my proof looks different than the one my prof posted as the answer; it's also way shorter. He used the one from the book but I didn't memorize it so I used this. Is it right?

Homework Statement



Prove that Absolute Convergence implies convergence


The Attempt at a Solution



Notation:

1. I'm using the notation an means a series. [As in: an = a(1)+a(2)+a(3)...+a(n) and n goes to infinity]

2. |an| means the absolute value of |an|.

***

I. |(-1)^n an| = an.

II. Suppose an converges. i.e: an = C

III. an converges implies that a(n+1) < a(n)

IV. (III.) implies (-1)^n an converges.

QED.
 
Last edited:

Answers and Replies

  • #2
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,473
255


There are a lot of problems with this.

(1) Your notation is inconsistent. In item (I), do you mean an or a(n)?

(2) (II) is false as stated. You are missing a "[itex]\lim_{n\rightarrow\infty}[/itex]".

(3) (III) is false. The terms of a series do not have to be monotonically decreasing in order for the series to converge. (They do have to converge to zero, though.)

(4) In (IV) you seem to be assuming that the series is alternating. This is one special case. But absolute convergence implies convergence for ANY series, not just alternating series. Also once again I assume you mean a(n) instead of an. By the way, a more standard notation would be [itex]s(n) = a(1) + \ldots + a(n)[/itex].

I recommend carefully reading the proof posted by your prof.
 
  • #3
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413


In (I), you're using an to mean two different things, and you have already confused me enough to not understand what you mean. In (II), you're making an assumption that appears to be irrelevant to the problem. Did you leave out some details? (III) is wrong. (IV) I'm not sure what you mean, but this looks wrong too.
 
  • #4


Yeah, as I'm reading this I see I should have clarified the following, which I assumed needn't be clarified:

(1) Your notation is inconsistent. In item (I), do you mean an or a(n)?

I mean an. As in, the absolute value of an alternating series is the same as the value of that series if it were non-alternating. (because an alternating series is essentially (-1)^n an.)

(2) (II) is false as stated. You are missing a "lim --> inf.".

Why? -- an is the sum of the series in this notation as n goes to infinity. That's what my definition says.

(3) (III) is false. The terms of a series do not have to be monotonically decreasing in order for the series to converge. (They do have to converge to zero, though.)

This was an actual mistake. I meant to write |a(n+1)| < |a(n)|

I also did not explain that if III is not satisfied by a series that goes negative (for example, 1/1, -1/2, 0, 0, 1/3, 0, -1/4; or a series involving something like sin(n), which goes negative and positive but is not an alternating series, then the proof still holds because such a series can always be expressed as smaller than a series that does satisfy III. For example, the one involving sin(n) can be expressed as a smaller convergent series because |sin(n)| ≤ 1)

I guess I should have included all those observations in my proof?

(IV) I'm not sure what you mean, but this looks wrong too.

IV simply uses the definition of a convergent alternating series (that if a(n + 1) < a(n), it converges).

Damn, was it really not easy to follow? it was so clear in my mind lol -- Other than my broad assumptions, did I make any actual mistakes? Given my explanation, is the proof valid?
 
Last edited:
  • #5


O, and I used proper sigma notation on my test. I just used weird notation here on PF to make things easier for myself on the computer (I'm not very comfortable with laTex).

So try and disregard the weird notation.
 
  • #6
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413


an alternating series is essentially (-1)^n an.
That doesn't actually make sense, because it would mean [tex](-1)^n\sum_{k=1}^\infty a_k[/itex], but I understand that you mean [tex]\sum_{k=1}^\infty(-1)^k a_k[/tex].

This was an actual mistake. I meant to write |a(n+1)| < |a(n)|
This still isn't true. For example, |a571| doesn't have to be smaller than |a570|.

Tip: Click the quote button to see how I typed the above.
 
  • #7


That doesn't actually make sense, because it would mean [tex](-1)^n\sum_{k=1}^\infty a_k[/itex], but I understand that you mean [tex]\sum_{k=1}^\infty(-1)^k a_k[/tex].

lol yes.

This still isn't true. For example, |a571| doesn't have to be smaller than |a570|.

That's what I meant. If the series doesn't have that property, like one that involves sin(n), such a series can still be expressed as one that is smaller than or equal to one that does have that property, or one that is convergent; and so it itself must be convergent.

For example, 1/2, 1/2, -1/3, -1/3, ... can be expressed as twice the alternating series 1/2, -1/3, ...

Or the series 1/2, 1/2, -1/3, -1/4, ... can be expressed as the series 1/2, -1/3... + the series 1/2, -1/4...

And the series sin(n)/n^p ≤ 1/n^p .

etc.

That's probably the missing step I should have added to seal the proof... unless you still think it doesn't hold?

It's what I had on my mind when I wrote it, but we only had 10 minutes per question, and it didn't occur to me to clarify that. It's the mindreader syndrome, lol; you think that because it's what you have in mind, the person reading it will see it that way.
 
Last edited:
  • #8
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,473
255


I mean an. As in, the absolute value of an alternating series is the same as the value of that series if it were non-alternating. (because an alternating series is essentially (-1)^n an.)

So you are asserting that

[tex]\left|\sum_{n=1}^{\infty}(-1)^n a_n\right| = \sum_{n=1}^{\infty} a_n[/tex]

or what? Because that is certainly not true.


Why? -- an is the sum of the series in this notation as n goes to infinity. That's what my definition says.

Well, from what you wrote I assumed you meant the n'th partial sum of the series. If an is the limit of the partial sums, then it does not depend on n, so your notation is extremely bad.

I also did not explain that if III is not satisfied by a series that goes negative (for example, 1/1, -1/2, 0, 0, 1/3, 0, -1/4; or a series involving something like sin(n), which goes negative and positive but is not an alternating series, then the proof still holds because such a series can always be expressed as smaller than a series that does satisfy III. For example, the one involving sin(n) can be expressed as a smaller convergent series because |sin(n)| ≤ 1)

I assume you mean that there exists a monotonically decreasing subsequence

[tex]a_{n_k}[/tex]

Whether or not this is true (can you prove it?), you can't conclude that

[tex]\sum_{n}a_n[/tex]

converges simply because

[tex]\sum_{k}a_{n_k}[/tex]

converges.

By the way, neither

[tex]\sum_n |\sin(n)|[/tex]

nor

[tex]\sum_n \sin(n)[/tex]

converges. The terms don't even converge to 0.

Damn, was it really not easy to follow? it was so clear in my mind lol -- Other than my broad assumptions, did I make any actual mistakes? Given my explanation, is the proof valid?

No, I don't think it's valid at all.
 
  • #9


So you are asserting that ...

I'm talking about absolute convergence, which means:

[tex]\sum_{n=1}^{\infty}\|\(-1)^n a_n\| = \sum_{n=1}^{\infty} a_n[/tex]

Which is true. Isn't it?

(I have no idea how to write it properly on laTex and I'm tired and on a train. it's supposed to be ∑|(-1)^1 an| )



I assume you mean that there exists a monotonically decreasing subsequence
...

I think you misunderstand what I wrote.

See my previous comment:

If the series doesn't have that property, like one that involves sin(n), such a series can still be expressed as one that is smaller than or equal to one that does have that property, or one that is convergent; and so it itself must be convergent.

For example, 1/2, 1/2, -1/3, -1/3, ... can be expressed as twice the alternating series 1/2, -1/3, ...

Or the series 1/2, 1/2, -1/3, -1/4, ... can be expressed as the series 1/2, -1/3... + the series 1/2, -1/4...

And the series sin(n)/n^p ≤ 1/n^p .

I know sin(n) doesn't converge. But a series containing sin(n) may converge; that's all I was talking about. I gave a series containing sin(n) as an example of a non-alternating series that sometimes goes negative.
 
  • #10
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413


I'm talking about absolute convergence, which means:

[tex]\sum_{n=1}^{\infty}\|\(-1)^n a_n\| = \sum_{n=1}^{\infty} a_n[/tex]

Which is true. Isn't it?

(I have no idea how to write it properly on laTex and I'm tired and on a train. it's supposed to be ∑|(-1)^1 an| )
There are few too many backslashes in your LaTeX. The code for the left-hand side is \sum_{n=1}^\infty|(-1)^n a_n|, but absolute convergence means

[tex]\sum_{n=1}^\infty |a_n|<\infty[/tex]

not

[tex]\sum_{n=1}^{\infty}|(-1)^n a_n|=\sum_{n=1}^\infty a_n[/tex]

See my previous comment:
...
For example, 1/2, 1/2, -1/3, -1/3, ... can be expressed as twice the alternating series 1/2, -1/3, ...
You shouldn't be rearranging terms if you don't know exactly what you're doing. For example 1-1+1-1+1-1+1-1+...=0, right?, but if you rearrange the terms, 1+1-1+1+1-1+1+1-1+...=1+1+1+1+...=∞.

I think you should probably start over, and be more careful with the definitions and the notation this time.
 
  • #11


I think you should probably start over, and be more careful with the definitions and the notation this time.

lol yeah I think I should rewrite the whole thing from the start carefully. I might do that tomorrow. I just arrived home, I was really tired and after 5 hours sitting on a train and a day without sleep I don't really have the energy to be rigorous (or even legible apparently :rofl:)

Anyway, thanks!
 

Related Threads on Is my proof for Absolute convergence => Convergence right?

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
9
Views
7K
Replies
6
Views
1K
Replies
6
Views
2K
Replies
9
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
Top