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Calculus Level 3

∑ n = 2 ∞ 1 ( ln ⁡ n ) ln ⁡ ( ln ⁡ n ) \large \sum_{n=2}^\infty \dfrac1{ ( \ln n)^{\ln (\ln n) } }

Does the series above converge or diverge?

Hint : For x ≥ 2 x\geq 2 , ln ⁡ ( x ) < x \ln(x) < \sqrt x .

Diverges Converges

Harshvardhan Mehta by Harshvardhan Mehta , 23, India

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2 solutions

We know that, ( l n ( n ) ) l n ( l n ( n ) ) = e l n ( l n ( n ) ) l n ( l n ( n ) ) = e [ l n ( l n ( n ) ) ] 2 { \left( ln\left( n \right) \right) }^{ ln\left( ln\left( n \right) \right) }={ e }^{ { ln\left( ln\left( n \right) \right) }^{ ln\left( ln\left( n \right) \right) } }={ e }^{ { \left[ ln\left( ln\left( n \right) \right) \right] }^{ 2 } }

and according to the hint, l n ( l n ( n ) ) < l n ( n ) ln\left( ln\left( n \right) \right) <\sqrt { ln\left( n \right) }

1 e [ l n ( l n ( n ) ) ] 2 > 1 e l n ( n ) = 1 n \frac { 1 }{ { e }^{ { \left[ ln\left( ln\left( n \right) \right) \right] }^{ 2 } } } >\frac { 1 }{ { e }^{ ln\left( n \right) } } =\frac { 1 }{ n }

Since ∑ 2 ∞ 1 n \sum _{ 2 }^{ \infty }{ \frac { 1 }{ n } } diverges, so does ∑ 2 ∞ 1 ( l n ( n ) ) l n ( l n ( n ) ) \sum _{ 2 }^{ \infty }{ \frac { 1 }{ { \left( ln\left( n \right) \right) }^{ ln\left( ln\left( n \right) \right) } } } .

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Moderator note:

Good observation that ( ln ⁡ n ) ln ⁡ ln ⁡ n < n \left( \ln n \right) ^ { \ln \ln n } < n .

How can one prove the hint that you gave?

Jake Lai
Dec 24, 2015

∑ n = n 0 ∞ f ( n ) \displaystyle \sum_{n=n_0}^\infty f(n) converges iff ∫ n 0 ∞ f ( x ) d x \displaystyle \int_{n_0}^\infty f(x) \ dx converges. We shall consider f ( x ) = 1 ( ln ⁡ x ) ln ⁡ ln ⁡ x f(x) = \dfrac{1}{(\ln x)^{\ln \ln x}} .

∫ 3 ∞ 1 ( ln ⁡ x ) ln ⁡ ln ⁡ x d x = ∫ ln ⁡ 3 ∞ 1 u ln ⁡ u d u 1 / e u = ∫ ln ⁡ 3 ∞ e u − ln ⁡ 2 u d u ≥ ∫ ln ⁡ 3 ∞ e u − u / 2 d u ( u / 2 ≥ ln ⁡ 2 u when u > > 0 ) ≥ ∫ ln ⁡ 3 ∞ e u / 2 d u \begin{aligned} \int_3^\infty \frac{1}{(\ln x)^{\ln \ln x}} \ dx &= \int_{\ln 3}^\infty \frac{1}{u^{\ln u}} \ \frac{du}{1/e^u} \\ &= \int_{\ln 3}^\infty e^{u-\ln^2u} \ du \\ &\geq \int_{\ln 3}^\infty e^{u-u/2} \ du \qquad (u/2 \geq \ln^2u \text{ when } u >> 0) \\ &\geq \int_{\ln 3}^\infty e^{u/2} \ du \\ \end{aligned}

The integral diverges, and therefore so does the sum.

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The integral test can only be applied if you've shown that the integrand is continuous, positive and decreasing.

Pi Han Goh - 5 years, 5 months ago

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It is obvious f f is continuous.

Let x ≥ y ≥ 3 x \geq y \geq 3 . Then clearly ln ⁡ ln ⁡ x ≥ ln ⁡ ln ⁡ y \ln \ln x \geq \ln \ln y , and so [ ln ⁡ ln ⁡ x ] 2 ≥ [ ln ⁡ ln ⁡ y ] 2 [\ln \ln x]^2 \geq [\ln \ln y]^2 , which means that e − [ ln ⁡ ln ⁡ x ] 2 ≤ e − [ ln ⁡ ln ⁡ y ] 2 e^{-[\ln \ln x]^2} \leq e^{-[\ln \ln y]^2} . Since f ( z ) = e − [ ln ⁡ ln ⁡ z ] 2 f(z) = e^{-[\ln \ln z]^2} , and e x > 0 e^x > 0 for all real x, this shows that f f is both positive and decreasing.

Is this correct? Thanks for helping me ensure the rigour of my solution :)

Jake Lai - 5 years, 5 months ago

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Actually, you should start off with x ≥ y ≥ 3 x \geq y \geq 3 . Do you know why?

Calvin Lin Staff - 5 years, 5 months ago

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@Calvin Lin – Right, because it is possible that ln ⁡ ln ⁡ z \ln \ln z is negative between 2 and e.

Jake Lai - 5 years, 5 months ago

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@Jake Lai – Right, in particular, even though ln ⁡ ( ln ⁡ 2 ) < ln ⁡ ( ln ⁡ 3 ) \ln (\ln 2) < \ln (\ln 3 ) , we have [ ln ⁡ ( ln ⁡ 2 ) ] 2 > [ ln ⁡ ( ln ⁡ 3 ) ] 2 [ \ln (\ln 2) ]^2 > [ \ln (\ln 3) ] ^2 .

Calvin Lin Staff - 5 years, 5 months ago

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