468 Real Analysis
=(Fn+Fn− 1 )^2 +(Fn−Fn− 1 )^2 −(Fn− 2 )^2
=(Fn+ 1 )^2 +(Fn− 2 )^2 −(Fn− 2 )^2 =(Fn+ 1 )^2.With this the problem is solved.
(Mathematical Reflections, proposed by T. Andreescu)
313.The function|sinx|is periodic with periodπ. Hence
lim
n→∞
|sinπ√
n^2 +n+ 1 |=lim
n→∞
|sinπ(√
n^2 +n+ 1 −n)|.But
nlim→∞(√
n^2 +n+ 1 −n)=nlim→∞n^2 +n+ 1 −n^2
√
n^2 +n+ 1 +n=
1
2
.
It follows that the limit we are computing is equal to|sinπ 2 |, which is 1.
314.The limit is computed as follows:
lim
n→∞(
n
k)(
μ
n)k(
1 −
μ
n)n−k=lim
n→∞n!
k!(n−k)!⎛
⎜
⎝
μ
n
1 −μ
n⎞
⎟
⎠
k
(
1 −
μ
n)n=
1
k!lim
n→∞n(n− 1 )···(n−k+ 1 )
(
n
μ− 1
)k ·nlim→∞(
1 −
μ
n)μn·μ=
eμ
k!lim
n→∞nk−( 1 +···+(k− 1 ))nk−^1 +···+(− 1 )k−^1 (k− 1 )!
1
μkn
k−(k
1) 1
μk−^1 nk− (^1) +···+(− 1 )k
=
1
eμ·k!·
1
1
μk=
μk
eμ·k!.
Remark.This limit is applied in probability theory in the following context. Consider a
large populationnin which an event occurs with very low probabilityp. The probability
that the event occurs exactlyktimes in that population is given by the binomial formula
P(k)=(
n
k)
pk( 1 −p)n−k.But fornlarge, the number( 1 −p)n−kis impossible to compute. In that situation we set
μ=np(the mean occurrence in that population), and approximate the probability by
the Poisson distribution