Suppose xn is a sequence of real numbers satisfying |xn- xn-1|≤ 2^-n for each n ≥ 2. Prove that the sequence is convergent.
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Have you learned about Cauchy sequences? – Eric Brier Mar 09 '24 at 16:12
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1"|xn- xn-1|", that's equal to $1$, which is not $\le 2^{-n}$, so there's no such sequence. – peterwhy Mar 09 '24 at 16:20
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Do you mean $|x_n-x_{n-1}|$? – Тyma Gaidash Mar 09 '24 at 16:21
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Because convergence in the real numbers is the same as the sequence being Cauchy, you can prove it is Cauchy by induction on the difference between $m$ and $n$ for arbitrary $m,n>N.$ If you need help understanding Cauchy sequences this is a relevant question.
(Hint: If you WLOG assume $m>n$ you can create a triangle inequality chain to find $|x_m-x_n|$ as a geometric sum, then you will be able to find such an N.)
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