Question

let x_1= 8 and x_n+1 = .5x_n +2 for all n in N. Show that (x_n) is bounded and monotone. Find the limit.

Answer 1

Just giving a few of the terms
\[x_{1} = 8 , x_{2} = 6, x_{3} = 5, x_{4} = 4{1\over2}, x_{5} = 4{1\over4},x_{6}=4{1\over8}\]
At this point it should become apparent that the function is approaching a limit of 4.

If x were 4, we would no longer have a monotonically decreasing function
\[x_{n} = 4, x_{n+1}=4,x_{n+2}=4\]

The original sequence is bounded below due to the fact that it has a lower limit and it is monotonically decreasing since
\[x_{n+1} < x_{n}\] for all n in N

Answer 2

cool. but not rigorous enough for my prof. But its a great start. Based on what you said I should: 1, show its bounded. 2. show its decreasing 3. now we now with montonically bounded sequences lim x_n exists.
3. x = lim x_n = lim x_n+1 so solve x = .5 x + 2

Answer 3

Yea, I figured this would be for an analysis course. Sorry I can't provide the rigorous proof that usually is required

Answer 4

but you gave me the outline. thanks. got it .