Proving Divergence of Sum of Fractions

In summary, the conversation is discussing how to prove that if \sum a_n > 0 and s_n = a_1 + ... + a_n and \sum a_n diverges, then \sum \frac{a_n}{1+a_n} diverges. The participants suggest using the comparison test and the Cauchy criterion, and discuss the importance of finding a bound for a_n. They also briefly mention the recent change in the edit function on the forum.
  • #1
ehrenfest
2,020
1

Homework Statement


Suppose [itex]a_n > 0[/itex], [itex]s_n =a_1 + ... + a_n[/itex], and [itex]\sum a_n[/itex] diverges,

a) Prove that

[tex]\sum \frac{a_n}{1+a_n}[/tex]

diverges.


Homework Equations





The Attempt at a Solution


Comparison with a_n fails miserably.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
is there a typo? i haven't thought about the problem yet but i notice s_n is nowhere in the problem
 
  • #3
ircdan said:
is there a typo? i haven't thought about the problem yet but i notice s_n is nowhere in the problem

it is in part b which I have not posted (yet).
 
  • #4
ehrenfest said:
it is in part b which I have not posted (yet).

ok cool
 
  • #5
there must be a mistake. if [itex] \sum a_i [/itex] diverges then you cannot prove that [itex] \sum \frac{a_n}{1+a_n} [/itex] converges. are you to show that the second series diverges? or is the first series supposed to converge?
 
  • #6
Grr. I am two for two today. I am supposed to prove the second series diverges. Sorry. I checked that like three times. I don't know what is wrong with me.

Anyone know why the edit button is missing from my opening post and in fact my second post above also?
 
Last edited:
  • #7
ehrenfest said:
Anyone know why the edit button is missing from my opening post and in fact my second post above also?

You can only edit 30 minutes after you post.
 
  • #8
The edit post button disappears within 24hrs.

I will do the edit to correct the initial post.
 
  • #9
Integral said:
The edit post button disappears within 24hrs.

I began this thread less than 8 hours ago as you can see from the time stamps.

Is it 24 hours or 30 minutes? I remember it definitely used to be longer than 30 minutes. If that has changed I would recommend changing it back.
 
Last edited:
  • #10
ahh typo, no wonder!

recall from rudin, sum(a_k) converges iff for all e > 0 there is N s.t. for all m > n > N |sum(a_k, k = n, ..., m)| < e

i'll be extra terse on purpose because this is a good problem, made me think

a_k > 0 by hypothesis(this is used in the proof of course)
claim sum(a_k/(1 + a_k)) converges => sum(a_k) converges

"proof sketch"
since sum(a_k/(1 + a_k)) converges, there is a number 0 < p < 1 and an integer N such that for all k > N, a_k < p.

we can pick an M s.t. for all m > n > M,

|sum(a_k, k = n, ..., m)| = |a_n + ... + a_m| = |(1 + a_n)a_n/(1 + a_n) + ... + (1 + a_m)a_m/(1 + a_m)| < (1 + p)|a_n/(1 + a_n) + ... + a_m/(1 + a_m)| < and now it's obviousgoodluck!
 
Last edited:
  • #11
ehrenfest said:
I began this thread less than 8 hours ago as you can see from the time stamps.

Is it 24 hours or 30 minutes? I remember it definitely used to be longer than 30 minutes. If that has changed I would recommend changing it back.

The time was changed recently due to users abusing the function. See the following thread.

https://www.physicsforums.com/showthread.php?t=229578
 
  • #12
ircdan said:
ahh typo, no wonder!

recall from rudin, sum(a_k) converges iff for all e > 0 there is N s.t. for all m > n > N |sum(a_k, k = n, ..., m)| < e

i'll be extra terse on purpose because this is a good problem, made me think

a_k > 0 by hypothesis(this is used in the proof of course)
claim sum(a_k/(1 + a_k)) converges => sum(a_k) converges

"proof sketch"
since sum(a_k/(1 + a_k)) converges, there is a number 0 < p < 1 and an integer N such that for all k > N, a_k < p.

we can pick an M s.t. for all m > n > M,

|sum(a_k, k = n, ..., m)| = |a_n + ... + a_m| = |(1 + a_n)a_n/(1 + a_n) + ... + (1 + a_m)a_m/(1 + a_m)| < (1 + p)|a_n/(1 + a_n) + ... + a_m/(1 + a_m)| < and now it's obviousgoodluck!

I do not know how you are getting your p but I understand your idea. Here is a complete proof.

If the series [tex]\sum \frac{a_n}{1+a_n}[/tex] converges, then [itex]a_n[/itex] must be bounded. Otherwise, given any [itex]\epsilon>0[/tex] and natural number N, I could find [itex]n \geq N[/itex] so that [tex]\frac{a_n}{1+a_n}[/tex] is between [itex]1-\epsilon[/itex] and [itex]\epsilon[/itex]. So, the terms of that series would not go to 0. So, let B bound [itex]a_n[/itex].

Then fix \epsilon \geq 0. Use the assumed convergence of [tex]\sum \frac{a_n}{1+a_n}[/tex] and the Cauchy criterion to find N so large that n,m \geq N implies that [tex]\sum_{k=n}^m \frac{a_k}{1+a_k} \leq \epsilon/(1+B)[/tex]. Then we have that [tex]\sum_{k=n}^m a_k = \sum_{k=n}^m a_k (1+a_k)/(1+a_k) \leq (1+B) \sum_{k=n}^m a_k/(1+a_k) < \epsilon[/tex]

So convergence of the second series implies convergence of the first.

Please confirm that this is correct.
 
  • #13
i don't understand your argument for bounding a_n, this doesn't mean it's wrong though
everything else makes sense to me

i did this, sum(a_k/(1 + a_k)) converges, so lim a_k/(1 + a_k) = 0, so there is an N s.t. for all k > N, a_k/(1 + a_k) < 1/2, so a_k < 1/2(1 + a_k) = 1/2 + a_k/2, so a_k/2 < 1/2, ie, a_k < 1 for all k > N, so that's enough, the whole p thing is just because this implies there is a 0 < p < 1 s.t. a_k < p for all k > N(sometimes this is very useful so I just pulled it out incase I would need it later), but yea it's not needed, all you need is a bound
 
Last edited:
  • #14
ircdan said:
i don't understand your argument for bounding a_n, this doesn't mean it's wrong though
everything else makes sense to me

In short, I was just saying that if we assume \sum \frac{a_n}{a_n+1} converges then \frac{a_n}{a_n+1} must go to zero and then its pretty clear that a_n must be bounded. Does that make sense? There are lots of ways to show that formally. Yes, all we need is a bound and we are just confusing each other getting a bound in different ways. :) Problem solved.
 
  • #15
part d:

What can be said about

[tex]\sum \frac{a_n}{1+n a_n}[/tex]

and

[tex]\sum \frac{a_n}{1+n^2 a_n}[/tex]
.

When a_n = 1, the first one converges and the second one diverges. I cannot prove that that always happens though.
 
  • #16
ehrenfest said:
part d:

What can be said about

[tex]\sum \frac{a_n}{1+n a_n}[/tex]

and

[tex]\sum \frac{a_n}{1+n^2 a_n}[/tex]
.

When a_n = 1, the first one converges and the second one diverges. I cannot prove that that always happens though.

i'm not sure what you mean about a_n = 1, if this is the case then sum(a_n) = sum(1) diverges and so the does the first one i think since it becomes sum(1/(1 + n))some observations, assuming a_n > 0 always I take it
if sum(a_n) converges, then they both converge by comparison since |a_n/(1 + na_n)| <= |a_n| and |a_n/(1 + n^2a_n| <= |a_n|for the first, note if a_n = 1/n, then the first and sum(a_n) both diverge, if a_n = 1/n^2 then sum(a_n) and the first both converge, so my gut feeling is that for the first, we have a sum(a_n) converges iff the first converges situationfor the second, things don't look as nice, note setting a_n = 1/n^2 then sum(a_n) and the second both converge. on the other hand, setting a_n = 1/n, we have that sum(a_n) diverges yet the second converges! so I think you can't say too much here, i'd look at more examplesi'd focus on the first and try to prove that if it converges, then so does sum(a_n), I think it's true but don't have time to work on it atm
 
Last edited:
  • #17
ircdan said:
i'm not sure what you mean about a_n = 1, if this is the case then sum(a_n) = sum(1) diverges and so the does the first one i think since it becomes sum(1/(1 + n))

Sorry, I meant that the first one diverges and the second one converges. Sorry.
 

What is the concept of proving divergence of sum of fractions?

The concept of proving divergence of sum of fractions is to show that the infinite sum of fractions does not converge to a finite value, but instead diverges to infinity or negative infinity.

How do you prove the divergence of sum of fractions using the limit comparison test?

The limit comparison test states that if the limit of the ratio of two sequences is a non-zero constant, then both sequences either converge or diverge together. So, to prove the divergence of sum of fractions, we can compare it to a known divergent sequence, such as the harmonic series.

Can you prove the divergence of sum of fractions using the integral test?

Yes, the integral test can also be used to prove the divergence of a series. If the integral of the series diverges, then the series itself must also diverge.

What is the difference between absolute and conditional convergence?

Absolute convergence refers to a series that converges regardless of the order in which the terms are added. Conditional convergence occurs when the series only converges if the terms are added in a specific order.

Are there any other methods for proving the divergence of sum of fractions?

Yes, there are other techniques such as the ratio test, the root test, and the Cauchy condensation test that can also be used to prove the divergence of a series. It is important to choose the most appropriate test for the given series.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
660
  • Calculus and Beyond Homework Help
Replies
1
Views
91
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
999
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
753
  • Calculus and Beyond Homework Help
Replies
1
Views
253
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top