Answer to Question #162729 in Calculus for Bholu

Question

Suppose {an} ∞ n=1 be a sequence of positive real numbers and 0 < x < 1. If an+1 < x · an for every n ∈ N, prove that limn→∞ an = 0.

Answer 1

First of all,

By applying the inequality multiple times

we find that $a_n < x^na_0, \forall n \in \mathbb{N}$ (as $a_n < x a_{n-1}<x^2a_{n-2}<...<x^n a_0$).

Now as $a_n \geq 0, \forall n\in \mathbb{N}$, and for any $a_0 \in \mathbb{R}^+$, $a_0x^n \underset{n\to\infty}{\to}0$, as $0<x<1$,

We see that $0 \leq a_n \leq a_0 x^n$, the right sequence converges to zero and thus we can conclude

that $a_n \underset{n\to\infty}{\to} 0$ as it is bounded by two sequences both converging to zero.