If |a_k+1-a_k|≤ 1/2^k for all k, then |a_m-a_n| n. What is the smallest N this bound guarantees for |a_m-a_n|<1100 with m,n≥ N?

Exercise · ZW-CAUCHY-SEQUENCE-007 · ★★☆ Intermediate

If \(|a_{k+1}-a_k|\le 1/2^k\) for all \(k\), then \(|a_m-a_n|<1/2^{n-1}\) for \(m>n\). What is the smallest \(N\) this bound guarantees for \(|a_m-a_n|<\frac1{100}\) with \(m,n\ge N\)?

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C. \(N=8\)

Explanation

We need \(2^{N-1}\ge100\). Since \(2^6=64<100\le128=2^7\), we take \(N-1=7\), i.e. \(N=8\). With \(N=7\) the bound is \(1/64>1/100\).

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Review the concept: Cauchy sequence

A sequence \((a_n)\) is Cauchy if, for every \(\varepsilon>0\), there is an index \(N\) beyond which any two terms are less than \(\varepsilon\) apart.

\[\forall\varepsilon>0\ \exists N:\ m,n\ge N\Rightarrow|a_m-a_n|<\varepsilon\]

If \(|a_{k+1}-a_k|\le 1/2^k\) for all \(k\), then \(|a_m-a_n|<1/2^{n-1}\) for \(m>n\). What is the smallest \(N\) this bound guarantees for \(|a_m-a_n|<\frac1{100}\) with \(m,n\ge N\)?

Explanation

Review the concept: Cauchy sequence

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