106 3 Real Analysis
Example.Prove that limn→∞n
√
n=1.Solution.The sequencexn= n
√
n−1 is clearly positive, so we only need to bound it
from above by a sequence converging to 0. For that we employ the binomial expansion
n=( 1 +xn)n= 1 +(
n
1)
xn+(
n
2)
xn^2 +···+(
n
n− 1)
xnn−^1 +xnn.Forgetting all terms but one, we can write
n>(
n
2)
xn^2 ,which translates toxn<
√
2
n− 1 , forn≥2. The sequence√
2
n− 1 ,n≥2, converges to 0,
and hence by the squeezing principle,(xn)nitself converges to 0, as desired.
The third example was published by the Romanian mathematician T. Lalescu in 1901
in theMathematics Gazette, Bucharest.
Example.Prove that the sequencean= n+^1
√
(n+ 1 )!−n√
n!,n≥1, is convergent and
find its limit.
Solution.The solution we present belongs to M. ̧Tena. It uses Stirling’s formula
n!=√
2 πn(n
e)n
·eθn
12 n, with 0<θn< 1 ,which will be proved in Section 3.2.11. Taking thenth root and passing to the limit, we
obtain
lim
n→∞n
√nn!=e.We also deduce that
lim
n→∞n+ 1
√nn! =nlim→∞
n+ 1
n·
n
√nn!=e.Therefore,
nlim→∞(n+√ 1
(n+ 1 )!
√nn!)n
=nlim→∞(
n(n+ 1 )√
((n+ 1 )!)n
(n!)n+^1)n
=nlim→∞(
n(n+ 1 )√
(n+ 1 )n
n!)n=nlim→∞(
n+ 1√
n+ 1
√nn!)n
=nlim→∞(
n+ 1
√nn!)n+n 1