# Infinite series Σ 1/(ln(e^n+e^-n)) = Σ 1/ln(n)?

1. Oct 22, 2009

### utleysthrow

1. The problem statement, all variables and given/known data

Prove whether $$\sum \frac{1}{ln(e^{n}+e^{-n})}$$ converges or diverges

2. Relevant equations

3. The attempt at a solution

(second post today... sorry, I just want to make sure I'm getting this right)

Since $$e^{n}+e^{-n}$$ goes to infinity as n goes to infinity, could I say that $$\sum \frac{1}{ln(e^{n}+e^{-n})}$$ is like $$\sum \frac{1}{ln(n)}$$?

I know that $$\sum \frac{1}{ln(n)}$$ is definitely divergent because it is > 1/n and I can use the comparison test.

But could I argue that $$\sum \frac{1}{ln(e^{n}+e^{-n})}$$ and $$\sum \frac{1}{ln(n)}$$ are essentially the same?

2. Oct 22, 2009

### hamster143

No, you can't say that... for example n^2 goes to infinity as n goes to infinity, sum of 1/n diverges and sum of 1/n^2 converges.

But you can notice that $$e^n + e^{-n} < e^n + e^n$$ ...

3. Oct 22, 2009

### DPMachine

Right, so $$\frac{1}{ln(e^{n}+e^{-n})} > \frac{1}{ln(e^{n}+e^{n})} = \frac{1}{ln(2)+ln(e^{n})} = \frac{1}{ln(2)+n}$$

$$\sum \frac{1}{ln(2)+n}$$ certainly looks like it diverges... but how would I prove that?

comparison test wouldn't work since it's not quite true that $$\sum \frac{1}{ln(2)+n} > 1/n$$

4. Oct 22, 2009

### Bohrok

If you could show that ∑ 1/ln(en + e-n) is greater than ∑ 1/ln(n) then you could compare them and show that the first one diverges. Otherwise, it won't help you.

Try another comparison with en + e-n < e2n

5. Oct 22, 2009

### hamster143

Does $$\sum \frac{1}{n+1} = 1/2 + 1/3 + 1/4 + ...$$ converge or diverge?

6. Oct 22, 2009

### Bohrok

n + 1 < 2n
1/(n + 1) > 1/(2n)

Σ 1/(2n) = (1/2)Σ 1/n diverges, so Σ 1/(n + 1) diverges.

7. Oct 22, 2009

### hamster143

Exactly, and also Σ 1/(n+1) = (Σ 1/n) - 1, another way to see that it diverges. And 1/(ln(2)+n) > 1/n+1.