Let \(0 < b < 1\). Show that the sequence $$ \begin{align*} b > b^2 > b^3 > b^4 > ... > 0, \end{align*} $$ converges to 0.


For the absolute value function definition and other properties see here.
For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Problem Discussion:

This was really tricky. We are given a sequence and to find its limit, the idea here is to extract a subsequence and then concluding that it also converges to the same limit as the original sequence by theorem 2.5.2 which says that subsequences of a convergent sequence converge to the same limit as the original sequence. But then we can also find the limit of the subsequence using a differen path. So’ll end up with two limits for the same subsequence. This is impossible since limits are unique and this will give us a direct us. This final answer is only the limit of the subsequence but also the limit of the sequence itself.

Proof:

Let \((b^n)\) is the sequence defined above. This sequence is decreasing and bounded below by definition. By the monotone convergence theorem, \((b^n)\) convergences to some number \(l\) such that \(b > l \geq 0\). To compute \(l\), let \((b^{2n})\) be a subsequence of \((b^n)\). The terms of \((b^{2n})\) are

$$ \begin{align*} b^2 > b^4 > b^6 > ... > 0. \end{align*} $$

Since \((b^{2n})\) is a subsequence of \((b^n)\), then by theorem 2.5.2, we must have that \((b^{2n}) \rightarrow l\).

But, note that \(b^{2n} = b^nb^n\). So we can use the algebraic limit theorem to conclude that \((b^{2n}) \rightarrow l*l = l^2\). So now we arrived at two results. First \((b^{2n}) \rightarrow l\) and now \((b^{2n}) \rightarrow l^2\). But we know that the limit of a sequence is unique from uniqueness of limits theorem (2.2.7), so we must have that \(l^2 = l\) and therefore \(l = 0\).

\(\blacksquare\)

References: