Test the series [ln(n)]^-1 for convergence.

[buffordboy23]

Homework Statement

Test the series $$\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)}$$ for convergence.

The attempt at a solution

My thoughts are a comparison test with the harmonic series. It's clear that 1/ln(n) > 1/n for $$n \geq 2$$. This implies that

$$\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=2} \frac{1}{n} = \left(\sum^{\infty}_{n=2} \frac{1}{n}\right) + 1 - 1 = \left(\sum^{\infty}_{n=1} \frac{1}{n}\right) - 1$$Is it sufficient to say that

$$\left(\frac{1}{ln\left(2\right)} - \frac{1}{2}\right) + \left(\frac{1}{ln\left(3\right)} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{ln\left(N\right)} - \frac{1}{N}\right) \geq 1$$

for some integer N (I think N=3 or 4...have to check again), so that

$$\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=1} \frac{1}{n}$$

and so the series must diverge. This makes sense but I wonder if there is a simpler way to express the divergence.

 

[Mark44]

Homework Statement

Test the series $$\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)}$$ for convergence.

The attempt at a solution

My thoughts are a comparison test with the harmonic series. It's clear that 1/ln(n) > 1/n for $$n \geq 2$$. This implies that

$$\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=2} \frac{1}{n} = \left(\sum^{\infty}_{n=2} \frac{1}{n}\right) + 1 - 1 = \left(\sum^{\infty}_{n=1} \frac{1}{n}\right) - 1$$

What you have above is all you need. The harmonic series diverges, and so does the same series with the first term removed. You have established that your series is term-be-term larger than the corresponding terms of the harmonic series (minus its first term), so your series diverges as well.

Is it sufficient to say that

$$\left(\frac{1}{ln\left(2\right)} - \frac{1}{2}\right) + \left(\frac{1}{ln\left(3\right)} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{ln\left(N\right)} - \frac{1}{N}\right) \geq 1$$

for some integer N (I think N=3 or 4...have to check again), so that

$$\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=1} \frac{1}{n}$$

and so the series must diverge. This makes sense but I wonder if there is a simpler way to express the divergence.

 

[buffordboy23]

What you have above is all you need. The harmonic series diverges, and so does the same series with the first term removed. You have established that your series is term-be-term larger than the corresponding terms of the harmonic series (minus its first term), so your series diverges as well.

Thanks for the feedback. What a trivial problem. I thought this was true, but it just didn't seem "strict" enough for proof.

So this suggests to me that the harmonic series is divergent for any starting index, say, n = 10^9, as n approaches infinity. This thought seems even more sketchy than the conventional n = 1 index on first appearance. I will work it out to see it, since it's been awhile since I did this stuff.

 

[Mark44]

So this suggests to me that the harmonic series is divergent for any starting index, say, n = 10^9, as n approaches infinity. This thought seems even more sketchy than the conventional n = 1 index on first appearance. I will work it out to see it, since it's been awhile since I did this stuff.

Yes, that's right. The thing is that although 10^9 seems pretty large, infinity is very much larger. The basic idea is that you can lop a finite number of terms from the beginning of a series without changing its basic behavior.