1. ## sequences and series

We define a sequence recursively in the following way:
s1=2 and sn=3+2sn−1 for n1
1. Show that sn is an increasing sequence,
2. Show that sn is bounded from above,
3. Compute limnsn

2. 1.3+(2sn)−1 < 3+2(sn+1)−1
3+2sn-1 < 3+2sn+1
increasing for all values n

2... i don't believe it is bounded from above,

A squence (Sn) is bounded above if there is a number M such that
Sn<= M for all n>=1 as this is continues to increase there's no M.

A squence (Sn) is bounded below if there is a number M such that
M<= Sn for all n>=1 this however is true. it is bounded below. M<2

3. limnsn of 3+2sn-1

3+2-1=

3. Originally Posted by Ashley87
We define a sequence recursively in the following way:

s1=2 and sn=3+2sn−1 for n1
1. Show that sn is an increasing sequence,
2. Show that sn is bounded from above,
3. Compute limnsn
You have $s_n = \sqrt{3 + 2 s_{n-1}}$.

1. and 2. should be OK (say so if that's not the case).

For 3. you know that a limit exists since the sequence is increasing and bounded above. Assume the limiting value is L. Then $s_n \rightarrow L \Rightarrow s_{n-1} \rightarrow L$. So $L = \sqrt{3 + 2L} \Rightarrow L^2 = 3 + 2L \, ....$

4. Originally Posted by steven_smith
A squence (Sn) is bounded above if there is a number M such that
Sn<= M for all n>=1 as this is continues to increase there's no M.
You are claiming an increasing sequence cannot be bounded above, unfortunatly this is not true. Not only is it not true in general it is not true in this case.

CB

5. Originally Posted by steven_smith
1.3+(2sn)−1 < 3+2(sn+1)−1
3+2sn-1 < 3+2sn+1
increasing for all values n
Due to limitations with his notation you have the wrong end of the stick about what the OP is asking.

The intended recurrence is that given in MrF's post.

CB

(If you continue responding inappropriatly to posts that you obviously have limmited understanding of I will start issuing infractions, and if I get fed up with that I will just ban you to save tha agravation)