5

As said above, does this series converge or diverge? Is there a certain identity/theorem to prove this?

$$\sum_{n=2}^{\infty }\frac{1}{n(\ln(n))^{2}}$$

DMcMor
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goldenlinx
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  • https://math.stackexchange.com/questions/2563446/on-convergence-of-bertrand-series-sum-limits-n-2-infty-frac1n-alpha – Guy Fsone Feb 10 '18 at 10:13

4 Answers4

8

Use integral test. Observe you have \begin{align} \int^\infty_3 \frac{dx}{x(\log x)^2} = \int^\infty_{\log 3}\frac{du}{u^2}<\infty. \end{align}

Jacky Chong
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6

The convergence of this series follows from the Kraft–McMillan inequality (Wikipedia link). There is a binary prefix code for the positive integers which encodes any integer $n\ge 2$ in $\lceil \log_2 n\rceil + 2 \lceil \log_2 \log_2 n\rceil + 1$ bits, as follows:

  • First, write down a sequence $11\dots1$ of length $\lceil \log_2 \log_2 n\rceil$, followed by $0$, communicating the value of $\lceil \log_2 \log_2 n\rceil$.
  • Then, write down $\lceil\log_2 n\rceil$ in binary, which takes $\lceil \log_2 \log_2 n\rceil$ bits, communicating the value of $\lceil\log_2 n\rceil$. (The previous step was necessary so we'd know when to stop reading the binary value.)
  • Then, write down $n$ in binary, which takes $\lceil \log_2 n\rceil$ bits. (The previous step was necessary so we'd know when to stop reading the binary value.)

Therefore, by the Kraft–McMillan inequality, $$\sum_{n=2}^\infty 2^{-(\lceil \log_2 n\rceil + 2 \lceil \log_2 \log_2 n\rceil + 1)} \le 1.$$ But $2^{-(\lceil \log_2 n\rceil + 2 \lceil \log_2 \log_2 n\rceil + 1)} \sim \frac{1}{2n (\log_2 n)^2}$, which is only off by a constant factor from $\frac{1}{n (\ln n)^2}$, so $$\sum_{n=2}^\infty \frac{1}{n (\ln n)^2}$$ must also converge.

Misha Lavrov
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  • Thank you for bringing the Kraft–McMillan inequality to our attention ;) (+1) – Olivier Oloa Apr 10 '17 at 05:56
  • I think it's actually very closely related to why this series converges! The sequence $\sum \frac1{n^2}, \sum \frac{1}{n(\ln n)^2}, \sum \frac{1}{n \ln n (\ln \ln n)^2}, \dots$ of ever more slowly converging series corresponds to a sequence of ever more efficient prefix codes for the integers. – Misha Lavrov Apr 10 '17 at 06:03
  • This is what I scroll through stackexchange for! – Bananach Apr 10 '17 at 06:22
5

If one knows the Cauchy condensation test, $$\sum_{n=1}^\infty a_n\text{ converges } \iff \sum_{n=1}^\infty 2^n a_{2^n}\text{ converges}$$ one may observe that $$ \sum_{n=2}^{\infty }\frac{2^n}{2^n(\ln(2^n))^{2}}=\frac1{\ln^2 2}\cdot\sum_{n=2}^{\infty }\frac{1}{n^{2}}<\infty $$ the given series is thus convergent.

Olivier Oloa
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0

This is a particular case of a Bertrand's series: $\;\displaystyle\sum_{n\ge 2}\frac1{n^\alpha\log^\beta n}$.

This series converges if and only if

  • $\alpha>1$ or
  • $\alpha=1$ and $\beta>1$.

This is easily proved by comparison to a Riemann series for the first case and by the integral test for the second case.

Bernard
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