1. ## Convergence of Sequences

This proof is in my notes. I would appreciate any guidance that anyone has to offer.

Let $\displaystyle (x_n)$ be a sequence of complex numbers. If there is a constant $\displaystyle c<1$ such that $\displaystyle |x_{n+1}-x_n|\leq c|x_n-x_{n-1}|$ for all $\displaystyle n\geq 2$, prove that $\displaystyle (x_n)$ is convergent.

Judging by where this is in my notes I'm assuming I might be supposed to use the definition of a Cauchy sequence to prove this.

2. Originally Posted by zebra2147
This proof is in my notes. I would appreciate any guidance that anyone has to offer.

Let $\displaystyle (x_n)$ be a sequence of complex numbers. If there is a constant $\displaystyle c<1$ such that $\displaystyle |x_{n+1}-x_n|\leq c|x_n-x_{n-1}|$ for all $\displaystyle n\geq 2$, prove that $\displaystyle (x_n)$ is convergent.

Judging by where this is in my notes I'm assuming I might be supposed to use the definition of a Cauchy sequence to prove this.

Hints:

1) I assume $\displaystyle 0<c<1$ , and prove now that $\displaystyle |x_{n+1}-x_1|<c^{n-1}|x_2-x_1|$

2) Now prove that $\displaystyle \forall k\in\mathbb{N}\,,\,\,|x_{n+k}-x_n}<p(c)|x_2-x_1|\,,\,\,with\,\, p(x)$ an integer polynomial

with zero free coefficient.

Tonio

3. Originally Posted by zebra2147
This proof is in my notes. I would appreciate any guidance that anyone has to offer.

Let $\displaystyle (x_n)$ be a sequence of complex numbers. If there is a constant $\displaystyle c<1$ such that $\displaystyle |x_{n+1}-x_n|\leq c|x_n-x_{n-1}|$ for all $\displaystyle n\geq 2$, prove that $\displaystyle (x_n)$ is convergent.

Judging by where this is in my notes I'm assuming I might be supposed to use the definition of a Cauchy sequence to prove this.
In You define $\displaystyle \Delta_{n} = x_{n+1} - x_{n}$ is ...

$\displaystyle \displaystyle x_{n} = x_{0} + \sum_{k=0}^{n} \Delta_{k}$ (1)

Now if there is a constant $\displaystyle c<1$ such that for n 'large onough' is $\displaystyle |\Delta_{n}|\le c\ |\Delta_{n-1}|$ , then for the ratio test the series $\displaystyle \displaystyle \sum_{n=0}^{\infty} \Delta_{n}$ converges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. I know I'm probably being stupid but how would you recommend proving that? Would you use the triangle inequality by starting with $\displaystyle |x_{n+1}-x_2|\geq 0$?

5. Originally Posted by zebra2147
I know I'm probably being stupid but how would you recommend proving that? Would you use the triangle inequality by starting with $\displaystyle |x_{n+1}-x_2|\geq 0$?
$\displaystyle |x_{n+3}-x_{n+2}|\le c|x_{n+2}-x_{n+1}|\le c^2|x_{n+1}-x_{n}|\le c^3|x_{n}-x_{n-1}|$

6. The Tonio's 1) can be shown by induction.

7. Plato...
I guess I don't really follow that... how do we know that $\displaystyle |x_{n+2}-x_{n+1}|<c|x_{n+1}-x_n|$ and so on.... Isn't this assuming that the sequence is strictly increasing or decreasing? Or am I confused?
I think if I understood this completely then a proof by induction would be relatively easy...

8. Originally Posted by zebra2147
how do we know that $\displaystyle |x_{n+2}-x_{n+1}|<c|x_{n+1}-x_n|$ and so on.... Isn't this assuming that the sequence is strictly increasing or decreasing? Or am I confused?
Absolutely not.
It is simply using the given.
From the given, we know $\displaystyle |x_3-x_2|\le c|x_2-x_1|$. RIGHT?

Step up one. $\displaystyle |x_4-x_3|\le c|x_3-x_2|$.
In that line replace $\displaystyle |x_3-x_2|$ to get $\displaystyle |x_4-x_3|\le c^2|x_2-x_1|$.

Step up one again. $\displaystyle |x_4-x_3|\le c^3|x_2-x_1|$.

Now prove by induction that
$\displaystyle |x_{n+1}-x_n|\le c^{n-1}|x_2-x_1|$