Advanced book on Mathematics Olympiad

3.1 Sequences and Series 105

(b)A sequence(xn)ntends to infinity if for every> 0 there existsn()such that for
n > n(),xn>.
The definition of convergence is extended toRn, and in general to any metric space,
by replacing the absolute value with the distance. The second method for finding the
limit is called the squeezing principle.

The squeezing principle.

(a) Ifan≤bn≤cnfor alln, and if(an)nand(cn)nconverge to the finite limitL, then
(bn)nalso converges toL.
(b)Ifan≤bnfor allnand if(an)ntends to infinity, then(bn)nalso tends to infinity.

Finally, the third method reduces the problem via algebraic operations to sequences
whose limits are known. We illustrate each method with an example. The first is from
P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics(Springer, 2004).

Example.Let(xn)nbe a sequence of real numbers such that

nlim→∞(^2 xn+^1 −xn)=L.

Prove that the sequence(xn)nconverges and its limit isL.

Solution.By hypothesis, for everythere isn()such that ifn≥n(), then

L−< 2 xn+ 1 −xn<L+.

For suchnand somek>0 let us add the inequalities

L−< 2 xn+ 1 −xn<L+,

2 (L−) < 4 xn+ 2 − 2 xn+ 1 < 2 (L+),

···

2 k−^1 (L−) < 2 kxn+k− 2 k−^1 xn+k− 1 < 2 k−^1 (L+).

We obtain

( 1 + 2 +···+ 2 k−^1 )(L−) < 2 kxn+k−xn<( 1 + 2 +···+ 2 k−^1 )(L+),

which after division by 2kbecomes
(
1 −

1

2 k

)

(L−)<xn+k−

1

2 k

xn<

(

1 −

1

2 k

)

(L+).

Now chooseksuch that| 21 kxn|<and| 21 k(L±)|<. Then form≥n+k,

L− 3 <xm<L+ 3 ,

and sincewas arbitrary, this implies that(xn)nconverges toL.