Advanced book on Mathematics Olympiad

154 3 Real Analysis

as

1

n

[

1

1 +^1 n

+

1

1 +^2 n

+···+

1

1 +nn

]

,

we recognize the Riemann sum of the functionf:[ 0 , 1 ]→R,f(x)= 1 +^1 xassociated to
the subdivisionx 0 = 0 <x 1 =n^1 <x 2 =^2 n<···<xn=nn=1, with the intermediate
pointsξi=ni∈[xi,xi+ 1 ]. It follows that

lim

n→∞

(

1

n+ 1

+

1

n+ 2

+···+

1

2 n

)

=

∫ 1

0

1

1 +x

=ln( 1 +x)

∣

∣^10 =ln 2,

and the problem is solved. 

We continue with a beautiful example from the book of G. Pólya, G. Szego, ̋ Aufgaben
und Lehrsätze aus der Analysis(Springer-Verlag, 1964).

Example.Denote byGnthe geometric mean of the binomial coefficients
(
n
0

)

,

(

n

1

)

,...,

(

n

n

)

.

Prove that

lim

n→∞

√nGn=√e.

Solution.We have
(
n
0

)(

n

1

)

···

(

n

n

)

=

∏n

k= 0

n!

k!(n−k)!

=

(n!)n+^1

( 1! 2 !···n!)^2

=

∏n

k= 1

(n+ 1 −k)n+^1 −^2 k=

∏n

k= 1

(

n+ 1 −k

n+ 1

)n+ 1 − 2 k

.

The last equality is explained by

∑n
k= 1 (n+^1 −^2 k)=0, which shows that the denominator
is just(n+ 1 )^0 =1. Therefore,

Gn= n+^1

√(

n

0

)(

n

1

)

···

(

n

n

)

=

∏n

k= 1

(

1 −

k

n+ 1

) 1 −n^2 +k 1

.

Taking the natural logarithm, we obtain

1

n

lnGn=

1

n

∑n

k= 1

(

1 −

2 k

n+ 1

)

ln

(

1 −

k

n+ 1

)

.