Advanced book on Mathematics Olympiad

470 Real Analysis

318.It is known that

lim

x→ 0 +

xx= 1.

Here is a short proof using L’Hôpital’s theorem:

lim

x→ 0 +

xx=lim

x→ 0 +

exlnx=elimx→^0 +xlnx=e

limx→ 0 +ln 1 x

x =elimx→ 0 +(−x)= 1.

Returning to the problem, fix>0, and chooseδ>0 such that for 0<x<δ,

∣∣

xx− 1

∣∣

<.

Then forn≥^1 δwe have
∣∣
∣∣
∣
n^2

∫ (^1) n
0
(xx+^1 −x)dx

∣∣

∣∣

∣

≤n^2

∫ (^1) n
0
|xx+^1 −x|dx,
=n^2
∫ (^1) n
0
x|xx− 1 |dx < n^2
∫ (^1) n
0
xdx=



2

.

It follows that

lim

n→∞

∫ (^1) n
0
(xx+^1 −x)dx= 0 ,
and so
nlim→∞n^2
∫ (^1) n
0
xx+^1 dx=nlim→∞n^2
∫ (^1) n
0
xdx=

1

2

.

(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
D. Andrica)

319.We will prove by induction onn≥1 that

xn+ 1 >

∑n

k= 1

kxk>a·n!,

from which it will follow that the limit is∞.
Forn=1, we havex 2 ≥ 3 x 1 >x 1 =a. Now suppose that the claim holds for all
values up throughn. Then

xn+ 2 ≥(n+ 3 )xn+ 1 −

∑n

k= 1

kxk=(n+ 1 )xn+ 1 + 2 xn+ 1 −

∑n

k= 1

kxk