Advanced book on Mathematics Olympiad

Real Analysis 489

=

1

x− 1

+

2 n+^2

1 −x^2 n+^2

.

This completes the inductive proof.
Because
1
x− 1
+nlim→∞

2 n+^1

1 −x^2 n+^1

=

1

x− 1

+mlim→∞

m

1 −xm

=

1

x− 1

,

our series converges to 1/(x− 1 ).
(C. Nast ̆ ̆asescu, C. Ni ̧ta, M. Brandiburu, D. Joi ̧ta, ̆ Exerci ̧tii ̧si Probleme de Algebra ̆
(Exercises and Problems in Algebra), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1983) ̆

353.The series clearly converges forx=1. We will show that it does not converge for
x =1.
The trick is to divide through byx−1 and compare to the harmonic series. By the
mean value theorem applied tof(t)=t^1 /n, for eachnthere existscnbetweenxand 1
such that
√nx− 1

x− 1

=

1

n

c

(^1) n− 1
.
It follows that
√nx− 1
x− 1

>

1

n

(max( 1 ,x))

(^1) n− 1

1

n

(max( 1 ,x))−^1.

Summing, we obtain

∑∞

n= 1

√nx− 1

x− 1

≥(max( 1 ,x))−^1

∑∞

n= 1

1

n

=∞,

which proves that the series diverges.
(G.T. Gilbert, M.I. Krusemeyer, L.C. Larson,The Wohascum County Problem Book,
MAA, 1996)

354.Using the AM–GM inequality we have

∑∞

n= 1

√

anan+ 1 ≤

∑∞

n= 1

an+an+ 1

2

=

1

2

∑∞

n= 1

an+

1

2

∑∞

n= 2

an<∞.

Therefore, the series converges.

355.There are exactly 8· 9 n−^1 n-digit numbers inS(the first digit can be chosen in 8
ways, and all others in 9 ways). The least of these numbers is 10n. We can therefore
write