# A question about numerical series

1. Aug 1, 2012

### nernstkizen

suppose an>0, $\sum$andiverges

then what can we say about $\sum$ an/1+nan

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

2. Aug 1, 2012

### 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?

3. Aug 1, 2012

### 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.

4. Aug 1, 2012

### chiro

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.

5. Aug 1, 2012

### SammyS

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

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 an as n → ∞ .

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

6. Aug 1, 2012

### nernstkizen

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 $\sum$ an diverges as n$\rightarrow$∞

Last edited: Aug 1, 2012
7. Aug 1, 2012

### SammyS

Staff Emeritus
Do you mean the sequence, an, diverges, or do you mean the series, $\sum$ an , diverges ?

8. Aug 1, 2012

### nernstkizen

Sorry, a typo.
I mean $\sum$ an diverges.

9. Aug 1, 2012

### uart

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.

Last edited: Aug 1, 2012
10. Aug 1, 2012

### 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.

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

11. Aug 1, 2012

### 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?

12. Aug 1, 2012

### nernstkizen

Thank you so much!
I get it :tongue:

13. Aug 2, 2012

### uart

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.