Question

Let x be any real number with −1 < x < 1. Then (x^n) n∈N is a

convergent sequence with limit 0.
Prove this using the Bernoulli Inequality

Answer 1

Using the Bernoulli Inequality, we proved that the sequence (x^n) for −1 < x < 1 converges to 0 as n tends to infinity, since |x|, raised to increasing powers, will produce a number that gets progressively smaller and approaches 0.

Step-by-step explanation:

The question concerns the convergence of the sequence (x^n) where n is a natural number, and x is a real number such that −1 < x < 1. We will prove that this sequence converges to 0 using the Bernoulli Inequality, which states that (1 + x)^n ≥ 1 + nx for any real number x ≥ -1.

Let's fix x to satisfy -1 < x < 0 since if x = 0, the sequence trivially converges to 0. Now, because 0 < 1 + x < 2, raising it to any natural number n will yield a number between 0 and 1. According to the Bernoulli Inequality, we get (1 + x)^n ≥ 1 + nx, but since x is negative, 1 + nx < 1. Therefore, 0 < (1 + x)^n < 1, meaning (1 + x)^n approaches 0 as n increases.

Now, for x in the range -1 < x < 1, we consider the absolute value |x|, which is less than 1. Similarly, the sequence (|x|^n) will approach 0 as n increases. Since |x^n| = (|x|^n), the sequence (x^n) will also converge to 0.