Subsequences and Accumulation Values
A sequence that does not converge may still have converging subsequences.
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Definition 1 (Subsequence). Let \((a_n)_{n\in\mathbb{N}}\) be a sequence in \(\mathbb{K}\). Let \((n_k)_{k\in\mathbb{N}}\) be a strongly monotonically increasing sequence with \(n_k\in\mathbb{N}\) for all \(k\in\mathbb{N}\). Then \((a_{n_k})_{k\in\mathbb{N}}\) is called a subsequence.
Example 2. Consider the sequence \((a_n)_{n\in\mathbb{N}}=(\frac1n)_{n\in\mathbb{N}}\). Then some subsequences are given by
\((a_{n_k})_{k\in\mathbb{N}}=(a_{2k})_{k\in\mathbb{N}}=(\frac12,\frac14,\frac16,\frac18,\ldots)\);
\((a_{n_k})_{k\in\mathbb{N}}=(a_{k^2})_{k\in\mathbb{N}}=(1,\frac14,\frac19,\frac1{16},\frac1{25},\ldots)\);
\((a_{n_k})_{k\in\mathbb{N}}=(a_{2^k})_{k\in\mathbb{N}}=(\frac12,\frac14,\frac18,\frac1{16},\frac1{32},\ldots)\);
\((a_{n_k})_{k\in\mathbb{N}}=(a_{k!})_{k\in\mathbb{N}}=(1,\frac12,\frac16,\frac1{24},\frac1{120},\frac1{720},\ldots)\).
Theorem 3 (Convergence of subsequences). Let \((a_n)_{n\in\mathbb{N}}\) be a convergent sequence in \(\mathbb{K}\) with \(\lim_{n\to\infty}a_n=a\). Then all subsequences \((a_{n_k})_{k\in\mathbb{N}}\) of \((a_n)_{n\in\mathbb{N}}\) are also convergent with \[\lim_{k\to\infty}a_{n_k}=a.\]
Attention 4. The existence of a convergent subsequence \((a_{n_k})_{k\in\mathbb{N}}\) does in general not imply the convergence of \((a_n)_{n\in\mathbb{N}}\). For instance, consider \((a_n)_{n\in\mathbb{N}}=((-1)^n)_{n\in\mathbb{N}}\). Both subsequences \[\begin{aligned} (a_{2k})_{k\in\mathbb{N}}\,&=((-1)^{2k})_{k\in\mathbb{N}}&&=(1,1,1,1,\ldots)\\ (a_{2k+1})_{k\in\mathbb{N}}\,&=((-1)^{2k+1})_{k\in\mathbb{N}}\!\!\!\!\!\!&&=(-1,-1,-1,-1,\ldots) \end{aligned}\] are convergent though \((a_n)_{n\in\mathbb{N}}=((-1)^n)_{n\in\mathbb{N}}\) is divergent.
However, we can “rescue” this statement by additionally claiming that \((a_n)_{n\in\mathbb{N}}\) is monotonic.
Theorem 5 (Subsequences of monotonic sequences). Let \((a_n)_{n\in\mathbb{N}}\) be a sequence in \(\mathbb{R}\). If \((a_n)_{n\in\mathbb{N}}\) is monotonic and there exists a convergent subsequence \((a_{n_k})_{k\in\mathbb{N}}\), then \((a_n)_{n\in\mathbb{N}}\) is convergent with \[\lim_{n\to\infty}a_n=\lim_{k\to\infty}a_{n_k}.\]
Proof: Denote \(a=\lim_{k\to\infty}a_{n_k}\). We just
consider the case where \((a_n)_{n\in\mathbb{N}}\) is monotonically
increasing (the remaining part can be done analogously to the
argumentations at the end of the proof of the Theorem about bounded and
monotonic sequences. Since \((a_{n_k})_{k\in\mathbb{N}}\) is also
monotonically increasing, we have that \(a=\sup\{a_{n_k}\,:\,k\in\mathbb{N}\}\).
Let \(\varepsilon>0\). Due to the
convergence and monotonicity of \((a_{n_k})_{k\in\mathbb{N}}\), there exists
some \(K\in\mathbb{N}\) such that for
all \(k\geq K\) holds \[a-\varepsilon<a_{n_k}\leq a.\] Now
assume that \(n\geq N=n_K\).
Monotonicity then implies that \(a-\varepsilon<a_{n_K}\leq a_n\leq a_{n_n}\leq
a\). In particular, we have that \[|a-
a_n|=a-a_n<\varepsilon.\]\(\Box\)
Definition 6 (Accumulation value). Let \((a_n)_{n\in\mathbb{N}}\) be a sequence in \(\mathbb{K}\). Then \(a\in \mathbb{K}\) is called accumulation value if there exists some subsequence \((a_{n_k})_{k\in\mathbb{N}}\) with \[a=\lim_{k\to\infty}a_{n_k}.\]
Attention 7 (Names). Accumulation values are often called by other names, like accumulation points, limits points or cluster points.
\(a\in \mathbb{K}\) is an accumulation value if and only if in every \(\varepsilon\)-neighbourhood of \(a\), there are infinitely many elements of the sequence \((a_{n})_{n\in\mathbb{N}}\).
Definition 8 (Accumulation values \(\pm\infty\)). A real sequence \((a_n)_{n\in\mathbb{N}}\) is said to have the (improper) accumulation value \(\infty\) if it is not bounded from above. Analogously, we define the (improper) accumulation value \(-\infty\) if it is not bounded from below.
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