Advanced book on Mathematics Olympiad

Real Analysis 483

=

24 / 4!

22 / 2!

=

1

3

.

We conclude that the original limit is

√

3.

(J. Dieudonné,Infinitesimal Calculus, Hermann, 1962, solution by Ch. Radoux)

345.Through a change of variable, we obtain

bn=

∫n

0 f(t)dt

n

=

xn

yn

,

wherexn=

∫n
0 f(t)dtandyn=n. We are in the hypothesis of the Cesàro–Stolz theorem,
since(yn)nis increasing and unbounded and

xn+ 1 −xn

yn+ 1 −yn

=

∫n+ 1

0 f(t)dt−

∫n

0 f(t)dt

(n+ 1 )−n

=

∫n+ 1

n

f(t)dt=

∫ 1

0

f(n+x)dx=an,

which converges. It follows that the sequence(bn)nconverges; moreover, its limit is the
same as that of(an)n.
(proposed by T. Andreescu for the W.L. Putnam Mathematics Competition)

346.The solution is similar to that of problem 342. BecauseP(x) >0, forx =
1 , 2 ,...,n, the geometric mean is well defined. We analyze the two sequences separately.
First, let

Sn,k= 1 + 2 k+ 3 k+···+nk.

Because

nlim→∞

Sn+ 1 ,k−Sn,k

(n+ 1 )k+^1 −nk+^1

=nlim→∞

(n+ 1 )k

(k+ 1

1

)

nk+

(k+ 1

2

)

nk−^1 +···+ 1

=

1

k+ 1

,

by the Cesàro–Stolz theorem we have that

nlim→∞

Sn,k

nk+^1

=

1

k+ 1

.

Writing

An=

P( 1 )+P( 2 )+···+P (n)

n

=am

Sn,m

n

+am− 1

Sn,m− 1

n

+···+am,

we obtain

lim

n→∞

An

nm

=

am

m+ 1

.