For the sequence defined by #a_"1"=2, a_"n+1"=1/(3-a_"n")#, how to show that it is bounded below by 0?

For the sequence defined by #a_"1"=2, a_"n+1"=1/(3-a_"n")#?

(a) Assuming that sequence is decreasing, show that it is bounded below by 0

(b) Explain why this means it must have a limit.

(c) Find the limit #lim_(nrarroo)a_"n"#

2 Answers
Mar 21, 2018

a) Let's start by writing the first few terms of the sequence.

#a_1 = 2#
#a_2 = 1/(3 - 2) = 1#
#a_3 = 1/(3 - 1) = 1/2#
#a_4 = 1/(3 - 1/2) = 2/5#
#a_5 = 1/(3 - 2/5) = 5/13#

As you can see, each term is getting smaller, but there is no way it's going to go below #0# because for this to happen, we would need #a_n > 3#, and since #a_(n + 1) < a_n#, and #a_1= 2#, #a_n# will never be greater than #3# therefore we can say that the sequence is bounded below by #0#.

b) Because this sequence converges to #0# as #x# approaches to infinity, we can say #lim_(n-> oo) a_n = 0#. This means that the sequence is maybe not divergent, but that the limit test is inconclusive (we would need another test to check for convergence).

c) Here's the formal proof that #lim_(n -> oo) a_n = 0#:

#lim_(n -> oo) a_n =(1/a_n)/(3/a_n - a_n/a_n) = 0/(-1) = 0#

As found above.

Hopefully this helps!

Mar 21, 2018

Answer:

See below.

Explanation:

Taking

#a_(k+1) = 1/(3-a_k)#
#a_k = 1/(3-a_(k-1))#

we have

#a_(k+1)-a_k = (a_k-a_(k-1))/((3-a_k)(3-a_(k-1))# and

#abs(a_(k+1)-a_k) = abs(a_k-a_(k-1))/abs((3-a_k)(3-a_(k-1))#

and for #0 le a_k < 2# we have

#abs(a_(k+1)-a_k) < abs(a_k-a_(k-1))#

then the sequence converges for

#0 le a_k < 2#

The convergence limit can be attained when

#a_oo = 1/(3-a_oo)# or

#3a_oo -a_oo^2-1=0# giving

#a_oo = 1/2(3pm sqrt5)# with the feasible

#a_oo = 1/2(3-sqrt5)#