Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 9F / ii

Solution

For any fixed

r > 0

lo g y = o (y^{1/ r})

y \to \infty

. For example, l'Hôpital's rule applied to

lo g y / y^{1/ r}

gives the limit zero. With

y = lo g n

, this implies

(lo g lo g n)^{r} < lo g n

eventually. Therefore

\frac{1}{n ( l o g l o g n ) ^{r}} > \frac{1}{n l o g n}

for all sufficiently large

n

. The comparison series diverges by the integral test for convergence, since

\int^{M} \frac{d x}{x l o g x} = lo g lo g M + constant ⟶ \infty.

n = 3 \sum \infty \frac{1}{n ( lo g lo g n ) ^{r}} diverges for every r > 0.

One may obtain the same growth bound by extending the given logarithmic estimate to real

y

through

⌈ y ⌉

and choosing any exponent smaller than

1/ r

; no special convergence test for iterated logarithms is needed.

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