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.

Janardhanan Sivaramakrishnan - 3 years, 4 months ago

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

Omar El Mokhtar - 3 years, 4 months ago

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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.

Janardhanan Sivaramakrishnan - 3 years, 4 months ago

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@Janardhanan Sivaramakrishnan 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 \(f\).

Oussama Boussif - 3 years, 4 months ago

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