Convergence of subsequences
Convergence of subsequences
All subsequences of a convergent sequence converge to the same limit as the original sequence.
Why is this useful? Two separate uses are listed on the reverse side.
Proof
This proof is just unwrapping then wrapping up the definition of convergence.
Proof. Let \( (a_n)_{n=1}^{\infty} \) be a convergence sequence, converging to \( L \in \mathbb{R} \). Proof by contradiction. Assume that there exists a subsequence \( (a_{f(n)})_{n=1}^{\infty} \), where \(f\) is an increasing function \( f : \mathbb{N} \to \mathbb{N} \), which doesn't converge to \( L \). So there exists a real \( \varepsilon > 0 \) and an \(N \in \mathbb{N} \) such that not all terms \( (a_{f(n)})_{n=N}^{\infty} \) are \( \varepsilon\)-close to \( L \). These terms are all terms within the original sequence, so \( (a_n)_{n=1}^{\infty} \) cannot converge to \( L \).
Surprisingly useful!
The theorem is extremely simple, and the proof too is trivial, yet the theorem is quite useful, almost as a 'trick' for thinking about sequences in a way that helps prove convergence/non-convergence.
Proving divergence
Utilize the power of Modus tollens (denying the consequent). Given an implication \( A \implies B \), proving \( \lnot B \) is sufficient to prove \( \lnot A \). In this way, subsequence convergence is a necessary criterion for a sequences convergence and finding a subsequence that doesn't converge is sufficient to demonstrate that a sequence doesn't converge.
Divergence criterion example
The sequence:
can be easily shown to be divergent as it has two subsequences which converge to different limits. One converges to \( \frac{1}{3} \):
While the other to \( -\frac{1}{3} \):
Proving convergence
If a convergent sequence contains subsequences which can be related to each other through the algebraic properties of limits, then we can use this relationship, along with the fact that the subsequences have the same limit, to deduce statements about the value of the limit.
Example
Consider the sequence: \( (b^n)_{n=1}^{\infty} \), where \( 0> b > 1 \). We know that:
If we consider the sequence \( (b^n b^n)_{n=1}^{\infty} = (b^{2n})_{n=1}^{\infty} \) , we can see that this sequence is a subsequence of \( (b^n)_{n=1}^{\infty} \), and thus, \( (b^{2n})_{n=1}^{\infty} \) and \( (b^{n})_{n=1}^{\infty} \) must have the same limit. Call this limit \( L \in \mathbb{R} \).
In addition, by the algebraic properties of limits, we know that \( \lim_{n \to \infty}b^{2n} = (\lim_{n \to \infty} b^n ) (\lim_{n \to \infty} b^n ) \). In other words, \( L = L^2 \). But the only real for which this is true is \( L = 0 \). So we have found that \( \lim_{n\to\infty}b^{n} = 0\).