# SEQUENCE

Prove that(x_n) is convergent

Note by Omar El Mokhtar
3 years, 4 months ago

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$$|x_{n+1}-x_n| \leq k |x_{n}-x_{n-1}|, |x_{n}-x_{n-1}| \leq k |x_{n-1}-x_{n-2}|$$ Combining, $$|x_{n+1}-x_n| \leq k^2 |x_{n-1}-x_{n-2}|$$

This would imply that $$|x_{n+1}-x_n| \leq k^n |x_{1}-x_{0}|$$.

Since, $$k \in [0,1]$$, $$k^n \to 0$$ as $$n \to \infty$$.

Hence, $$|x_{n+1}-x_n| \rightarrow 0$$ as $$n$$ tends to infinity.

The series is thus convergent.

- 3 years, 4 months ago

I did the same proof but my teacher told me it must be shown that lim(x_n)=a

- 3 years, 4 months ago

The condition that you have given is for asymptotic convergence. Without an additional condition of $$|x_{n+1}-a|\le k|x_n-a|$$ the limiting condition cannot be proved.

With my knowledge of dynamics, there seems to be no solution without apriori assumption that $$\lim_{n \to \infty} x_n$$ exists and is finite.

- 3 years, 4 months ago

That's right, without any other conditions the only inequation we need to prove That a sequence converges to some value is the one You provided, and that's how we usually solve for the limits. Furthermore, we use This frequently when we have a sequence $$u_{n}$$ satisfying: $$u_{n+1}=f(u_{n})$$ For some continuous and differentiable function $$f$$.

- 3 years, 4 months ago