Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 10F / v / Solution

Part (iii) and the given harmonic-sum asymptotic show that

\frac{E T _{n}}{n l o g n} = \frac{H _{n}}{l o g n} ⟶ 1.

For any fixed

ε > 0

, this deterministic ratio lies within

ε /2

of

1

for all sufficiently large

n

. The Chebyshev inequality and part (iv) then give

P (\frac{T _{n}}{n lo g n} - 1 > ε) \leq P (∣ T_{n} - E T_{n} ∣ > \frac{ε}{2} n lo g n) \leq \frac{4 var ( T _{n} )}{ε ^{2} n ^{2} ( lo g n ) ^{2}} \leq \frac{4 C}{ε ^{2} ( lo g n ) ^{2}} ⟶ 0.

Thus

\frac{T _{n}}{n lo g n} ⟶ 1 in probability.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version