SEQUENCE

Prove that(x_n) is convergent

Note by Omar El Mokhtar
6 years, 4 months ago

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Comments

xn+1xnkxnxn1,xnxn1kxn1xn2|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, xn+1xnk2xn1xn2|x_{n+1}-x_n| \leq k^2 |x_{n-1}-x_{n-2}|

This would imply that xn+1xnknx1x0|x_{n+1}-x_n| \leq k^n |x_{1}-x_{0}|.

Since, k[0,1]k \in [0,1], kn0k^n \to 0 as nn \to \infty.

Hence, xn+1xn0|x_{n+1}-x_n| \rightarrow 0 as nn tends to infinity.

The series is thus convergent.

Janardhanan Sivaramakrishnan - 6 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 - 6 years, 4 months ago

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The condition that you have given is for asymptotic convergence. Without an additional condition of xn+1akxna|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 limnxn\lim_{n \to \infty} x_n exists and is finite.

Janardhanan Sivaramakrishnan - 6 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 unu_{n} satisfying: un+1=f(un)u_{n+1}=f(u_{n}) For some continuous and differentiable function ff.

Oussama Boussif - 6 years, 4 months ago
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