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Test the convergence or divergence of this series whose $n^{\text{th}}$ term is: $$(\ln n)^{-\ln(\ln n)},$$ i.e. Test the convergence or divergence of this series: $$\sum_{i=1}^{\infty}(\ln n)^{-\ln(\ln n)}.$$

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Denote $$a_n = \frac{1}{\big(\ln n\big)^{\ln\ln n}}$$

Two closely related solutions:

  1. Integral Test:

$\sum_{n=2}^{\infty}a_n$ converge or diverges with $$\int_2^\infty \frac{dx}{\big(\ln x\big)^{\ln\ln x}} \overset{\substack{x\equiv e^u,\\dx = e^u\,du\\\vphantom{x}}}{=\mkern-3mu=\mkern-3mu=\mkern-3mu=} \int_{\ln 2}^{\infty} \frac{e^u \;du}{u^{\ln u}} = \int_{\ln 2}^{\infty} \frac{e^u}{e^{(\ln u)^2}}\;du = \int_{\ln 2}^{\infty} \underbrace{e^{u - (\ln u)^2}}_{\to\infty}\;du = \infty$$ so it diverges.

  1. Cauchy condensation test ($\sum a_n$ and $\sum 2^na_{2^n}$ converge or diverge together).

This is essentially the same argument. The sum convergence or divergence of the sum with terms

$$ b_n = 2^na_{2^n} = \frac{2^n}{\big( \ln 2^n \big)^{\ln\ln 2^n}} > \frac{2^n}{\big( \ln e^n \big)^{\ln\ln e^n}} = \frac{e^{n\ln 2}}{e^{(\ln n)^2}} = e^{n\ln(2) -(\ln n)^2} \to \infty $$ as the polynomial in the exponent dominates and goes $\to\infty$. So $\sum b_n$ diverges, so $\sum a_n$ diverges.

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  • $\begingroup$ Perfect one.. can it be done by cauchy? $\endgroup$ – Ünpredictable Kaushik Apr 15 '18 at 7:21
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By Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

we have

$$\sum_{i=1}^{\infty}\frac1{(\ln n)^{\ln(\ln n)}} \ge\frac12 \sum_{i=1}^{\infty}\frac{2^n}{(\ln 2^n)^{\ln(\ln 2^n)}} =\frac12\sum_{i=1}^{\infty}\frac{2^n}{(n\ln 2)^{{\ln(n\ln 2)}}}>\frac12\sum_{i=1}^{\infty}\frac{2^n}{n^{{\ln(n\ln 2)}}}$$

and

$$\frac{2^n}{n^{{\ln(n\ln 2)}}}\to \infty$$

indeed

$$a_n=\frac{2^n}{n^{{\ln(n\ln 2)}}}\implies \sqrt[n] a_n=\frac{2}{n^{\frac{{\ln(n\ln 2)}}{n}}}\to 2$$

since

$$n^{\frac{{\ln(n\ln 2)}}{n}}=e^{ \frac{{\ln(n\ln 2)}\cdot\ln n}{n} }\to e^0=1$$

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  • $\begingroup$ How can you say the sequence $\frac{2^n}{n^{{\ln(n\ln 2)}}}\to \infty $ $\endgroup$ – Ünpredictable Kaushik Apr 15 '18 at 7:15
  • $\begingroup$ @ÜnpredictableKaushik The best way is by root test. $\endgroup$ – user Apr 15 '18 at 7:17
  • $\begingroup$ I am not getting the limit value by root test? $\endgroup$ – Ünpredictable Kaushik Apr 15 '18 at 7:18
  • $\begingroup$ @ÜnpredictableKaushik I add this step here in a moment. $\endgroup$ – user Apr 15 '18 at 7:20
  • $\begingroup$ $$2^n / n^{\ln(n \ln 2)} = e^{n \ln(2)} / e^{\ln(n \ln 2) \cdot \ln n} = e^{n \ln(2) - \ln(n \ln 2) \cdot \ln n} $$ The exponent is then $$n \ln(2) - (\ln n)^2 - \ln 2 \cdot \ln n \to \infty$$ $\endgroup$ – adfriedman Apr 15 '18 at 7:21

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