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

[DPMachine]

Homework Statement

Given that $$a_{n} > 0$$ and $$lim(na_{n}) = l$$ with $$l\neq0$$,
prove that $$\sum a_{n}$$ diverges.

Homework Equations

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.

 

[Office_Shredder]

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

 

[DPMachine]

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

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?

 

[Office_Shredder]

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

This part is true

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?

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

 

[DPMachine]

This part is true
This part isn't. For example, if a_n=1/n, then na_n converges to 1, but a_n converges to 0!

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.

 

[Office_Shredder]

You have that -l/n<a_n<3l/n

This doesn't tell you anything! a_n 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

|na_n-l|<e, what can you say about |a_n-l/n|?

 

[DPMachine]

You have that -l/n<a_n<3l/n

This doesn't tell you anything! a_n 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

|na_n-l|<e, what can you say about |a_n-l/n|?

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

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

 

[Office_Shredder]

Stop trying to bound it below by a single number. You know a_n 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

|a_n-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 a<sub>n? 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?</sub>

 

[DPMachine]

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

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

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

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

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

 

[DPMachine]

Stop trying to bound it below by a single number. You know a_n 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

|a_n-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 a<sub>n? 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?</sub>

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

 

[Office_Shredder]

Probably not...

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