-1
$\begingroup$

Let $a_n$ be a sequence that fulfils

$|a_{n+1}-a_n|<\lambda\cdot|a_n-a_{n-1}|$

$\forall n\geq2$ and $0<\lambda<1$

Prove that $a_n$ converges.

  • 0
    Surely you mean a sequence not a series? – 2017-02-13
  • 0
    Try borrowing arguments from the proof of [Contraction Mapping Principle](https://en.wikipedia.org/wiki/Banach_fixed-point_theorem). – 2017-02-13
  • 0
    Try using recursion. – 2017-02-13
  • 0
    The distance between the terms seems to be getting smaller. I suggest cauchy? – 2017-02-13
  • 0
    Missing from the problem statement is the nature of the sequence $\{a_n\}$. Are these real numbers? Complex numbers? I assume one or the other is true, judging by the tags, but it would improve the Question to say so in the body of the Question. – 2017-02-14
  • 0
    Not specified in the original question. But obviously it's R – 2017-02-16

2 Answers 2

1

Phrased differently, we have

$$0 \leq d(a_{n+1},a_n)< \lambda^n d(a_1-a_0) \implies 0 \leq d(a_{n+k},a_n)<\sum_{i=n}^{m}\lambda^{i} d(a_1-a_0)$$

and since the geometric series converges, we know that the rightmost sum gets arbitrarily small as $n \to \infty$, so the sequence is cauchy, and hence convergent in a complete metric space ($\mathbb R$)

2

By induction $b_n:=|a_{n+1}-a_n| < \lambda^n|a_1-a_0|$ and since $0<\lambda<1$ we have that $\{b_n\}_n$ is bounded by a convergent geometric series.