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

[utleysthrow]

Homework Statement

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

Homework Equations

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?

 

[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 ...

 

[DPMachine]

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 ...

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

 

[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

 

[hamster143]

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

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

 

[Bohrok]

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

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

 

[hamster143]

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

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

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.