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
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1compare it with harmonic series – jimjim Apr 25 '16 at 21:57
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What about sum of $\frac{1}{n}$ – openspace Apr 25 '16 at 21:57
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Have tried it, but then I will need to prove that also $n\ge log^3n$ with induction maybe, which isn't that easy... – CryoDrakon Apr 25 '16 at 21:58
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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.
Sri-Amirthan Theivendran
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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.
Mark Viola
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Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola May 26 '16 at 17:09