Prove the divergent series $\sum_{k=2}^{\infty} \frac{1}{log^3k}$

Question

Prove that the series $$\sum_{k=2}^{\infty} \frac{1}{log^3k}$$ diverges. I have already tried the ratio test and root test but both give me that it's less than 1, but when I wanted to check it on Wolfram Alpha it says it diverges by the comparison test. Sadly I cant find any function for camparing it to and that is where I am stuck

Answer 1

Use the cauchy condensation test. Note that $$2^na_{2^n}=\frac{2^n}{\log^3(2^n)}=\frac{2^n}{n^3\log^32}$$

so that $\sum 2^n a_{2^n}$ does not converge (terms don't go to zero) and hence the original series diverges.

Answer 2

In THIS ANSWER, I showed using only the limit definition of the logarithm and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\log(x)\le x-1 <x \tag 1$$

for $x>0$. Now, noting that $\log(x^{\alpha})=\alpha \log(x)$, we find using $(1)$ that for any $\alpha>0$

$$\log(n)\le \frac{n^\alpha}{\alpha} \tag 2$$

Finally, we find from $(2)$ that for any $\alpha>0$

$$\frac{1}{\log^3(n)}\ge \frac{\alpha^3}{n^{3\alpha}}$$

Setting $\alpha=1/3$ yields

$$\frac{1}{\log^3(n)}\ge \frac{1}{27 n}$$

from which the comparison theorem guarantees that the series of interest diverges.