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Does $ L = \sum_{i=2}^{\infty } \frac{1}{n \log(n)} $ converge or diverge?

I established that: $$ L \le I = \int^{\infty}_{2} \frac{1}{n \log(n)} = \lim_{n \to \infty} [ \ln(\log(n)) - \ln(\log(2)) ], $$ and as $ \lim_{n \to \infty} \log(x) = \infty $, then $ L $ diverges.

But I'm not sure:

  1. of the sense of the inequality,
  2. about the conclusion.
gt6989b
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freehumorist
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    Note that you proved that $I=\infty$, and if you start from $L \leq I$, you cannot say anything (because nothing is greater than $\infty$). However, the standard way of proof shows that actually $L \geq I$. Since $I=\infty$, then $L=\infty$. – Aphelli May 08 '19 at 18:47
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    You want to show that $L$ is greater than or equal to some integral that you can show is $+\infty$. Since $\frac{1}{x\log(x)}$ it is a decreasing function (on $(1,\infty)$ at least), it suffices to choose $L$ to be the left hand rectangle rule for a suitable integral. In fact it is the left hand rectangle rule for $I$, so in fact $L \geq I$ and $I=+\infty$ so $L=+\infty$. – Ian May 08 '19 at 18:56
  • @Ian thank you! it helped me to figure out the inequality. – freehumorist May 08 '19 at 18:58

1 Answers1

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What you want to use is the integral test: the series (with positive terms) $\;\sum_{n=2}^\infty\frac1{n\log n}$ converges (resp. diverges) if and only if the improper integral $\;\int_2^\infty\frac 1{x\log x}\,\mathrm dx$ converges (resp. diverges).

Now this integral is easy to calculate by substitution: setting $u=\log t$, we obtain $$\int_2^A \frac 1{x\log x}\,\mathrm dx=\int_{\log 2}^{\log A} \frac{\mathrm d u}u= \log\, \biggl|_{\log 2}^{\log A}=\log(\log A)-\log(\log 2)\xrightarrow[A\to\mkern1mu+\infty]{}+\infty.$$

Bernard
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