Advanced book on Mathematics Olympiad

Real Analysis 481

The conclusion follows.

341.In view of the Cesàro–Stolz theorem, it suffices to prove the existence of and to
compute the limit

nlim→∞

(n+ 1 )p

(n+ 1 )p+^1 −np+^1

.

We invert the fraction and compute instead

lim

n→∞

(n+ 1 )p+^1 −np+^1

(n+ 1 )p

.

Dividing both the numerator and denominator by(n+ 1 )p+^1 , we obtain

nlim→∞

1 −

(

1 −n+^11

)p+ 1

1

n+ 1

,

which, with the notationh=n^1 + 1 andf(x)=( 1 −x)p+^1 , becomes

−lim

h→ 0

f (h)−f( 0 )

h

=−f′( 0 )=p+ 1.

We conclude that the required limit isp+^11.

342.An inductive argument shows that 0<xn<1 for alln. Also,xn+ 1 =xn−xn^2 <xn,
so(xn)nis decreasing. Being bounded and monotonic, the sequence converges; letxbe
its limit. Passing to the limit in the defining relation, we find thatx=x−x^2 ,sox=0.
We now apply the Cesàro–Stolz theorem. We have

nlim→∞nxn=nlim→∞

n

1

xn

=nlim→∞

n+ 1 −n

1

xn+ 1 −

1

xn

=nlim→∞

1

1

xn−xn^2 −

1

xn

= lim

n→∞

xn−xn^2

1 −( 1 −xn)

=lim

n→∞

( 1 −xn)= 1 ,

and we are done.

343.It is not difficult to see that limn→∞xn=0. Because of this fact,

nlim→∞

xn

sinxn

= 1.

If we are able to find the limit of

n

1

sin^2 xn

,