# Understanding a statement about the series $S =\sum_2 ^\infty \frac{1}{n\ln n}$

In Mathematical Methods for Physicists, the author writes the following with reference to the series $$S =\sum_2 ^\infty \frac{1}{n\ln n}$$:

We form the integral $$\int_2^\infty \frac{1}{x\ln x} dx$$ which diverges, indicating that $$S$$ is divergent. Because $$n\ln n \gt n$$, the divergence is slower than that of the harmonic series.

So far so good. He further writes:

But because $$\ln(n)$$ increases more slowly than $$n^\epsilon$$, where $$\epsilon$$ can have an arbitrarily small positive value, we have divergence even though the series $$\sum_n n^{-{(1+\epsilon)}}$$ converges.

This is the part that confuses me.
Perhaps the author is referring to the comparison test. It is known that $$\sum_n n^{-{(1+\epsilon)}}$$, i.e. $$\sum_n \frac{1}{n.n^\epsilon}$$ converges since $$1+\epsilon \gt 1$$. It would suffice to prove that $$n\ln(n) \gt n.n^\epsilon$$. For this we compare how fast $$\ln(n)$$ and $$n^\epsilon$$ increase with $$n$$. Let's take their derivatives, $$\frac{1}{n}$$ and $$\frac{\epsilon}{n^{1-\epsilon}}$$. Now we do not know the nature of $$\epsilon$$. How do I proceed with the conclusion?

• The author presented several statements: (i) $S$ diverges by comparing with $\int_2^\infty \frac{1}{x\ln x} dx$; (ii) $S$ grows slower than harmonic series; (iii) $S$ grows faster than $\sum_n n^{-{(1+\epsilon)}}$; (iv) $\sum_n n^{-{(1+\epsilon)}}$ converges. There is no contradiction (as expected), and (iii) (iv) don't imply (i) directly. Example: $\sum_2 ^\infty \frac{1}{n(\ln n)^2}$. Commented Dec 22, 2021 at 15:38
• There are many other convergence tests. The Cauchy Condensation Test works for $\sum_n 1/(n\log n)$. Commented Dec 23, 2021 at 8:34

## 2 Answers

By $$n\ln n < n^{1+\epsilon}$$, thus $$\sum \dfrac{1}{n\ln n} > \sum \dfrac{1}{n^{1+\epsilon}}$$. The series $$\sum \dfrac{1}{n^{1+\epsilon}}$$ converges. This cannot be conclusive about the convergence of $$S$$, because $$S$$ is bigger than something finite then we don't know if $$S$$ is finite or infinite. Hence, you should use integral.

• How did you conclude that $nlnn \lt n^{1+\epsilon}$ without knowing the nature of $\epsilon$? Commented Dec 23, 2021 at 3:47
• Formally they meant that for every fixed $\epsilon > 0$ there is an $n_0$ such that for all $n > n_0$ we have $n \ln(n) < n^{1 + \epsilon}$. Commented Dec 23, 2021 at 13:02

To avoid all confusions and sticking to explanations using words like "faster" and "slower" which reduce the rigour of the argument you should directly look into the Integral Test or the Cauchy Condensation test.

By Cauchy Condensation test it suffices to check convergence of the series $$\sum \frac{2^{n}}{2^{n}n\ln(2)}=\sum \frac{1}{n\ln(2)}$$ which diverges as $$\sum\frac{1}{n}$$ is divergent.

• It cannot be true that $\frac 1{n\ln n} \geq \frac 1 n$. To prove that $\sum \frac 1{n \ln n}$ diverges, it is not enough just to compare it with $\sum \frac 1 n$. You need to compare it with an integral. Commented Dec 22, 2021 at 20:07
• Thanks for answering.All my confusion stemmed from the existence of $\epsilon$ in the equation. Could you please edit your answer accordingly? Commented Dec 23, 2021 at 3:51