prove that $a_n$ converges while $n\ge2$: $|a_{n+1}-a_n|<c|a_n-a_{n-1}|$

Question

Given that $a_n$ fulfills following condition: there is a real number $0<c<1$ so for all natural $n\ge2$: $|a_{n+1}-a_n|<c|a_n-a_{n-1}|$.

Prove that $a_n$ converges.

I need to prove it by using Cauchy sequence.

I tried to prove it by defining $a_n$ at first, but I can't figure out how to evaluate: $|a_{n+p}-a_n|$.

Answer 1

HINT: $|a_{n+k+1}-a_{n+k}|<c^k|a_n-a_{n-1}|$. Can you find a bound for $|a_{n+k+m}-a_{n+k}|$?

Answer 2

Hint: $$|a_{n+p}-a_n| \le \sum_{k = 1}^p |a_{n+k}-a_{n+k-1}| \le \sum_{k = 1}^p c^k|a_{n}-a_{n-1}| \le \sum_{k = 1}^\infty c^k|a_{n}-a_{n-1}|.$$ Do you know how to continue?

The sum of the geometric-series $\sum\limits_{k = 1}^\infty c^k$ is $\dfrac{1}{1-c}$, and $|a_n-a_{n-1}| < c^{n-1} |a_2-a_1|$, so $$|a_{n+p}-a_n| \le \frac{1}{1-c} \, c^{n-1}|a_2-a_1| \xrightarrow[n\to\infty]{} 0.$$ Hence $(a_n)$ is Cauchy.

Answer 3

From the imperial m1p1 past papers? Hint: Write $|a_m - a_n| = |(a_m - a_{m-1}) + (a_{m-1} - a_{m-2}) + ....| $ and apply triangle inequality, you can then substitute in your original expression and obtain a sum which will give you the desired results.