A question about numerical series

[nernstkizen]

suppose a_n>0, \suma_ndiverges

then what can we say about \sum a_n/1+na_n

I think it divergent, but I'm not sure。

 

[chiro]

Hey nernstkizen and welcome to the forums.

Have you looked at tests for convergence (like ratio test) to get some ideas of how to possibly show this? What have you thought about or tried?

 

[nernstkizen]

hi~

I think it is impossible to use any test to solve it directly, maybe some tricks should be introduced.

btw, it is the exercise 3.11(d) of Principles of Mathematical Analysis (Rudin). Some solutions said that it can either converge or diverge, but I don't agree with them.

 

[chiro]

hi~

I think it is impossible to use any test to solve it directly, maybe some tricks should be introduced.

btw, it is the exercise 3.11(d) of Principles of Mathematical Analysis (Rudin). Some solutions said that it can either converge or diverge, but I don't agree with them.

The thing I was thinking of is to use the facts about series that diverge.

For example if a series diverges then by the ratio test (assuming all positive numbers), the ratio between successive terms (an+1/an) must be greater than 1. Similarly you get other kinds of results that you can use.

The idea is to get as many of these results as possible and then incorporate them into your new series: you know that an+1/an > 1 in the ratio test so if you can re-arrange your ratio test in the second way to get this term then you can replace with some variable which is greater than 1.

 

[SammyS]

suppose a_n>0, \suma_n diverges .

then what can we say about \sum a_n/1+na_n ?

I think it's divergent, but I'm not sure。

Do you mean \displaystyle \sum \ \frac{a_n}{1+n\,a_n}\ ? That could be written as \sum a_n/(1+na_n) .

What you wrote literally means \displaystyle \sum \ \frac{a_n}{1}+n\,a_n\ .


As to whether \displaystyle \sum \ \frac{a_n}{1+n\,a_n}\ diverges or converges, that may depend upon the behavior of a_n as n → ∞ .

You can rewrite your series as \displaystyle \sum \ \frac{1}{(1/a_n)+n}\ .

 

[nernstkizen]

Do you mean \displaystyle \sum \ \frac{a_n}{1+n\,a_n}\ ? That could be written as \sum a_n/(1+na_n) .

What you wrote literally means \displaystyle \sum \ \frac{a_n}{1}+n\,a_n\ .As to whether \displaystyle \sum \ \frac{a_n}{1+n\,a_n}\ diverges or converges, that may depend upon the behavior of a_n as n → ∞ .

You can rewrite your series as \displaystyle \sum \ \frac{1}{(1/a_n)+n}\ .

Thanks for your reply !

1 What I mean is \sum \frac{a_n}{1+na_n}

2 I have already transformed it to \displaystyle \sum \ \frac{a_n}{1+n\,a_n}\ ,
and I think it won't converges as long as \sum a_n diverges as n\rightarrow∞

 

[SammyS]

Thanks for your reply !

1 What I mean is \sum \frac{a_n}{1+na_n}

2 I have already transformed it to \displaystyle \sum \ \frac{a_n}{1+n\,a_n}\ ,
and I think it won't converges as long as a_n diverges as n\rightarrow∞

Do you mean the sequence, a_n, diverges, or do you mean the series, \sum a_n , diverges ?

 

[nernstkizen]

Do you mean the sequence, a_n, diverges, or do you mean the series, \sum a_n , diverges ?

Sorry, a typo.
I mean \sum a_n diverges.

 

[uart]

btw, it is the exercise 3.11(d) of Principles of Mathematical Analysis (Rudin). Some solutions said that it can either converge or diverge, but I don't agree with them.

I'm far from an expert on series nernstkizen, but I tend to think you're correct that it always diverges. My reasoning is as follows.

Consider the limiting value of just the n \, a_n part of the expression (as n goes to infinity). It may either,

1. go to zero.
2. remain bounded (say bounded by M).
3. go to infinity.

In case one, the tail of the new series approaches that of the original. So the series diverges.

In case two, the terms in the tail of the new series are bounded below by a constant multiple, 1/(1+M), of the original series. So again it diverges.

And in the third case, the terms of the tail of the series approaches 1/n. So again it diverges.

 

[voko]

This series can both converge and diverge. Taking a_{n} = \frac {1} {n}, both series diverge.

Now take a_{n} = 1, if n = 2^{q} - 1 and a_{n} = 0, otherwise.

\sum a_{n} diverges because it sums infinitely many 1's.

\sum \frac {a_{n}}{1 + na_{n}} = \sum 0 + \sum \frac {1}{2^{q}} = 2

 

[voko]

Oops, I missed the requirement a_{n} > 0. But the proof can be trivially modified to account for that and have the series converge. Can you see how?

 

[nernstkizen]

Oops, I missed the requirement a_{n} > 0. But the proof can be trivially modified to account for that and have the series converge. Can you see how?

Thank you so much!
I get it

 

[uart]

This series can both converge and diverge. Taking a_{n} = \frac {1} {n}, both series diverge.

Now take a_{n} = 1, if n = 2^{q} - 1 and a_{n} = 0, otherwise.

Good example, thanks for posting that voko.

I was having trouble seeing how it could converge, as I was looking at it with "blinkers" on and only thinking about series that were monotonic in the tail.