- #1

- 532

- 2

**Homework Statement**

Test the series [tex]\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} [/tex] 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 [tex] n \geq 2 [/tex]. This implies that

[tex]\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[/tex]

Is it sufficient to say that

[tex] \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 [/tex]

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

[tex]\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=1} \frac{1}{n} [/tex]

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

Last edited: