# Divergence of an infinite series (using the def of limit)

1. Oct 23, 2009

### DPMachine

1. The problem statement, all variables and given/known data
Given that $$a_{n} > 0$$ and $$lim(na_{n}) = l$$ with $$l\neq0$$,
prove that $$\sum a_{n}$$ diverges.

2. Relevant equations

3. The attempt at a solution

lim(na_n)=l (with =/= 0), so I can safely say that:

$$\left|na_{n}-l\right| < \epsilon$$ by the definition of limit.

Then isn't it also true that $$\left|a_{n}-l\right| < \epsilon$$ because $$\left|a_{n}-l\right| \leq \left|na_{n}-l\right|$$ and is smaller than the same epsilon?

From there it would imply that a_n converges to l which is never 0, so the sum of a_n would not converge either.

Last edited: Oct 24, 2009
2. Oct 24, 2009

### Office_Shredder

Staff Emeritus
In fact, you expect |an-l| to be approximately l for large values of n. Why? Because nan converges to a value, so an must go to zero since n goes to infinity

3. Oct 24, 2009

### DPMachine

I made a mistake on the original post. The goal is to prove that $$\sum a_{n}$$ diverges. Sorry about that... Does my explanation make more sense now?

4. Oct 24, 2009

### Office_Shredder

Staff Emeritus
This part is true

This part isn't. For example, if an=1/n, then nan converges to 1, but an converges to 0!

5. Oct 24, 2009

### DPMachine

ok, thank you. I think I proved it by letting epsilon = 2l (which is >0 because a_n >0)

Then $$-2l < na_{n}-l < 2l$$ implies $$-l < na_{n} < 3l$$, which implies $$-l/n < a_{n}$$

Since ∑ -l/n = (-1)*∑ 1/n which doesn't converge, by comparison test a_n doesn't converge.

6. Oct 24, 2009

### Office_Shredder

Staff Emeritus
You have that -l/n<an<3l/n

This doesn't tell you anything! an could be, for example, 1/n2 and satisfy this. The comparison test only works when all the terms in both series are positive

When you have

|nan-l|<e, what can you say about |an-l/n|?

7. Oct 24, 2009

### DPMachine

|an-l/n| approaches 0 doesn't it? I'm not sure if this was what you were getting at... I understand that an is some form of 1/n, but I want to prove that by using the e def of limit and without providing a specific an that works.

How about if e = l/2? Then I would have l/2 < an < l/2 + l. Couldn't I use the comparison test on l/2 < an? Thanks again.

8. Oct 24, 2009

### Office_Shredder

Staff Emeritus
Stop trying to bound it below by a single number. You know an goes to zero, so if you can bound it below by l/2, you have a contradiction. I don't see how you got that last inequality

|an-l/n| goes to zero, but how?

Look:
$$|na_n-l|< \epsilon \rightarrow |a_n-\frac{l}{n}|< \frac{ \epsilon}{n}$$

So what can we say about an? For large enough n, this:

$$\frac{l}{n} - \frac{ \epsilon}{n} < a_n < \frac{l}{n} + \frac{ \epsilon}{n}$$

So what value of epsilon do you want, and what series are you comparing to?

9. Oct 24, 2009

### DPMachine

If $$\epsilon = \frac{l}{2}$$, we have $$\left| na_{n} - l \right| < \epsilon = \frac{l}{2}$$

which $$= -\frac{l}{2} < na_{n} - l < \frac{l}{2}$$

$$= -\frac{l}{2n} < a_{n} - l < \frac{l}{2n}$$

$$= -\frac{l}{2n} + l < a_{n} < \frac{l}{2n} + l$$

$$= \frac{l}{2n} < a_{n} < \frac{l}{2n} + l$$

10. Oct 24, 2009

### DPMachine

Can $$\epsilon = \frac{l+\epsilon}{n}$$?...

11. Oct 24, 2009

### Office_Shredder

Staff Emeritus
Probably not...

You had the right idea with comparison test. You just need to figure out what value of epsilon gives you something useful