Key to "Prove ∑{n=2}∞1nln⁡(n)ln⁡ln⁡n is divergent."

Vikolers6

Answered question

2022-01-21

Prove

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

is divergent.

Answer & Explanation

sirpsta3u

Beginner2022-01-21Added 42 answers

Your series diverges by the integral test, because

\int \frac{1}{x \log (x) \log (\log (x))} d x = \log (\log (\log (x)))

and because

lim_{x \to + \infty} \log (\log (\log (x))) = + \infty

Concerning your approach, note that the inequality

\frac{1}{n \ln^{2} n} < \frac{1}{n^{2}}

is false. Actually,

n \ln^{2} n < n^{2}

poleglit3

Beginner2022-01-22Added 32 answers

HINT
Let use Cauchy condensation test
$0 \leq \sum_{{n = 1}}^{\infty} f (n) \leq \sum_{{n = 0}}^{\infty} 2^{n} f (2^{n}) \leq 2 \sum_{{n = 1}}^{\infty} f (n)$
$\sum_{{n = 2}}^{\infty} \frac{1}{n \ln (n) \ln \ln n} \geq \frac{1}{2} \sum_{{n = 2}}^{\infty} \frac{2^{n}}{2^{n} \ln (2^{n}) {\ln \ln 2}^{n}} = \frac{1}{2}$
$\sum_{{n = 2}}^{\infty} \frac{1}{n \ln 2 \ln (n \ln 2)}$
and
$\sum_{{n = 2}}^{\infty} \frac{1}{n \ln 2 \ln (n \ln 2)} \geq \frac{1}{2} \sum_{{n = 2}}^{\infty} \frac{2^{n}}{2^{n} \ln 2 \ln (2^{n} \ln 2)} = \frac{1}{2}$
$\sum_{n = 2}^{\infty} \frac{1}{\ln 2 \cdot (n \ln 2 + \ln \ln 2)}$

RizerMix

Expert2022-01-27Added 656 answers

Alternatively, use Ermakoffs

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