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Erschienen in:
2015 | OriginalPaper | Buchkapitel
5. Infinite Sequences II
verfasst von : Miklós Laczkovich, Vera T. Sós
Erschienen in: Real Analysis
Verlag: Springer New York
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Abstract
Finding the limit of a sequence is generally a difficult task. Sometimes, just determining whether a sequence has a limit is tough. Consider the sequence (18) in Example 4.1, that is, let a
n
be the nth digit in the decimal expansion of \(\sqrt{2}\). We know that (a
n
) does not have a limit. But does the sequence \(c_{n} = \root{n}\of{a_{n}}\) have a limit? First of all, let us note that a
n
≥ 1, and thus c
n
≥ 1 for infinitely many n. Now if there are infinitely many zeros among the terms a
n
, then c
n
= 0 also holds for infinitely many n, so the sequence (c
n
) is divergent. However, if there are only finitely many zeros among the terms a
n
, that is, a
n
≠ 0 for all n > n
0, then 1 ≤ a
n
≤ 9, and so \(1 \leq c_{n} \leq \root{n}\of{9}\) also holds if n > n
0. By Theorem 4.17, \(\root{n}\of{9} \rightarrow 1\). Thus for a given ɛ > 0, there is an n
1 such that \(\root{n}\of{9} <1+\varepsilon\) for all n > n
1. So if \(n>\max (n_{0},n_{1})\), then \(1 \leq c_{n} <1+\varepsilon\), and thus c
n
→ 1.