# Divergent series

1. Nov 10, 2005

### happyg1

Hello, I'm working on this problem:
Prove that for any real x, the series SUM n=2 to infinity of 1/(log n)^x diverges.

So far, I have applied the test that says that if SUM 2^n*a_2n converges then the series converges. I got:

1/log2*SUM 2^n/n^x I know that 1/n^x converges if x>1, but 2^n will explode, so I'd have convergent (sometimes) times divergent = divergent. Can I do that?

Thanks,
CC

2. Nov 10, 2005

### quasar987

No, that is not a propery of series. A counter exemple is

$$\sum \frac{1}{n}\frac{1}{n^2}$$

In this case, we have divergent times convergeant = convergeant.

What you can do to prove the desired result however, is simply show that for all x in R, the limit as n-->infty of the argument of the serie is not 0. Hence it does not converges, hence it diverges. (You'll have to do the 3 separate cases x<0, x=0 and x>0 [the later being the most difficult]).

3. Nov 10, 2005

### happyg1

I tried it using the ratio test. I got
$$\2(\frac{n}{n+1})^x[\tex] and the limit of that is 2>1 so it diverges. can I do it that way? CC 4. Nov 10, 2005 ### happyg1 let me try again...I'm still learning latex.... [tex]\mbox{2}(\frac{n}{n+1})^x$$
there. And the limit of THAT is 2.
Is this correct?

5. Nov 10, 2005

### NateTG

Yes.
$$\lim_{n \rightarrow \infty} 2 \left( \frac{n}{n+1}\right)^x=2[/itex] But [tex] \frac{(\log n)^x}{(\log n+1)^x} \neq 2 \left( \frac{n}{n+1}\right)^x$$

Last edited: Nov 10, 2005
6. Nov 10, 2005

### quasar987

You should do it my way. It's much more instructive then mechanically using a test anyway.