3.1 Sequences and Series 107=
(
nlim→∞n+ 1
√nn!)limn→∞nn+ 1
=e.Taking thenth root and passing to the limit, we obtainnlim→∞n+√ (^1) (n+ 1 )!
√nn! = 1 ,
and hence
lim
n→∞
an
√nn!=nlim→∞
n+√ (^1) (n+ 1 )!
√nn! − 1 = 0.
Thus, if we set
bn=
(
1 +
an
√nn!)√nann!
,then limn→∞bn=e. From the equality
(n+√ 1
(n+ 1 )!
√nn!)n
=bann√nn!
n ,we obtain
an=ln(n+√ 1
(n+ 1 )!
√nn!)n
(lnbn)−^1(
n
√nn!)− 1
.
The right-hand side is a product of three sequences that converge, respectively, to 1=lne,
1 =lne, and^1 e. Therefore, the sequence(an)nconverges to the limit^1 e.
Apply these methods to the problems below.313.Compute
lim
n→∞∣
∣
∣sin(
π√
n^2 +n+ 1)∣∣
∣.
314.Letkbe a positive integer andμa positive real number. Prove that
nlim→∞(
n
k)(
μ
n)k(
1 −μ
n)n−k
=μk
eμ·k!.
315.Let(xn)nbe a sequence of positive integers such thatxxn=n^4 for alln≥1. Is it
true that limn→∞xn=∞?