Question

Suppose that a sequence (an) satisfies the condition |an+1 − an+2| < 1/2 |an − an+1|.
Prove that (an) converges.

Answer 1

i think this is hard, and may need induction
you want to show the sequence is a cauchy sequence

Answer 2

since \(|a_3-a_2|<\frac{1}{2}|a_2-a_1|\) we know \[|a_4-a_3|<\frac{1}{2}|a_3-a_2|<\frac{1}{2^2}|a_2-a_1|\]

Answer 3

by induction you get
\[|a_{n+2}-a_{n+1}|<\frac{1}{2^n}|a_2-a_1|\] which is getting closer to proving it is a cauchy sequence, but not quite there yet

Answer 4

perhaps you had an example in class where you had to show if
\[|a_{n+1}-a_n|<2^{-n}\] then the sequence was a cauchy sequence
you can mimic that here

Answer 5

What if I didn't

Answer 6

hmm then you have some work to do
depends on what you can use
it is a fact that if \(|a_{n+1}-a_n|<b_n\) where \(\sum b_n\) is finite, then \({a_n}\) is cauchy
but if you did not do this , then what you have to show is that
\[|a_n-a_m|\to 0\] for arbitrary \(m,n\)
in this case you will need the triangle inequality repeatedly

Answer 7

best bet is to google the proof that if \(|a_{n+1}-a_n|<\frac{1}{n^2}\) then \(\{a_n\}\) is cauchy

Answer 8

look at #4 here
http://math.stackexchange.com/questions/182830/proving-that-a-sequence-is-cauchy
more than i can write out, but it is all there

Answer 9

ok that was a typo, i meant google the proof that \(|a_{n+1}-a_n|<\frac{1}{2^n}\) etc
i reversed the \(n\) and \(2\) by mistake

Answer 10

Thank you kind sir. May God be with you in your travels