# 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"#

For the sequence defined by

(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

##### 2 Answers

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

b) Because this sequence converges to

c) Here's the formal proof that

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

As found above.

Hopefully this helps!

#### Answer:

See below.

#### Explanation:

Taking

we have

and for

then the sequence converges for

The convergence limit can be attained when