Advanced book on Mathematics Olympiad

3.2 Continuity, Derivatives, and Integrals 155

This is just a Riemann sum of the function( 1 − 2 x)ln( 1 −x)over the interval[ 0 , 1 ].
Passing to the limit, we obtain

nlim→∞

1

n

lnGn=

∫ 1

0

( 1 − 2 x)ln( 1 −x)dx.

The integral is computed by parts as follows:
∫ 1

0

( 1 − 2 x)ln( 1 −x)dx

= 2

∫ 1

0

( 1 −x)ln( 1 −x)dx−

∫ 1

0

ln( 1 −x)dx

=−( 1 −x)^2 ln( 1 −x)

∣

∣^1

0 −^2

∫ 1

0

( 1 −x)^2

2

·

1

1 −x

dx+( 1 −x)ln( 1 −x)

∣∣

∣∣

1

0

+x

∣

∣∣

∣

∣

1

0

=−

∫ 1

0

( 1 −x)dx+ 1 =

1

2

.

Exponentiating back, we obtain limn→∞n

√

Gn=

√

e. 

468.Compute

lim

n→∞

[

1

√

4 n^2 − 12

+

1

√

4 n^2 − 22

+···+

1

√

4 n^2 −n^2

]

.

469.Prove that forn≥1,

1

√

2 + 5 n

+

1

√

4 + 5 n

+

1

√

6 + 5 n

+···+

1

√

2 n+ 5 n

<

√

7 n−

√

5 n.

470.Compute

nlim→∞

(

21 /n

n+ 1

+

22 /n

n+^12

+···+

2 n/n

n+^1 n

)

.

471.Compute the integral
∫π

0

ln( 1 − 2 acosx+a^2 )dx.

472.Find all continuous functionsf:R→[ 1 ,∞)for which there exista∈Randka
positive integer such that

f(x)f( 2 x)···f (nx)≤ank,

for every real numberxand positive integern.