(last updated: 4:24:03 PM, September 27, 2020)

\(\large \S\) 2.2 - Limit theorems

Now that you are familiar with the concept of convergence for sequences, we’ll explore various theorems that will help you determine convergence when sequences are combined in various ways or how to deduce convergence or divergence of one sequence in comparison to another sequence which you know something about.

Combining sequences

Theorem 2.2.1 If sequences \(a_n\) and \(b_n\) converge to \(A\) and \(B\), respectively, then:
  1. \(\displaystyle\lim_{n\rightarrow\infty}(a_n\pm b_n)=A\pm B\).
  2. \(\displaystyle\lim_{n\rightarrow\infty}a_nb_n=AB\).
  3. \(\displaystyle\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\frac A B\), as long as \(B\neq0\) (and \(b_n\neq0\), otherwise \(\frac{a_n}{b_n}\) is undefined for some \(n\)).
  4. \(\displaystyle\lim_{n\rightarrow\infty}(a_n)^p=A^p\) for all \(p\in\mathbb N\).
  5. \(\displaystyle\lim_{n\rightarrow\infty} \sqrt[k]{a_n}=\sqrt[k]{A}\) for all \(k\in\mathbb N\) so long as \(a_n\geq0\) if \(k\) is even.
  6. If \(a_b\leq b_n\) for all \(n\geq n_1\) (for some \(n_1\in\mathbb N\)), then \(A\leq B\).

Proofs of (a), (b), (c), and (f) are in the textbook; (d) is homework; (e) is a very similar argument to (d). For proofs of (d) and (e), see my supplemental notes on exponentiation of sequences where I prove a slightly different result, but the method to prove (d) and (e) here is essentially identical.

Example

\(a_n=5+\frac1n\) converges to \(5\). Thus we know that \(b_n=\left(5+\frac1n\right)^{10}=(a_n)^{10}\) converges to \(5^{10}\) by Theorem 2.2.1(d) and that \(b_n=\sqrt{5+\frac1n}=\sqrt{a_n}\) converges to \(\sqrt{5}\) by Theorem 2.2.1(e).

Note that this theorem gives us that \(a_n\rightarrow A\) implies that \(a_n^p\rightarrow A^p\), but that the converse is not necessarily true.

Counterexample

\(a_n=(-1)^n\) does not converge, but \(a_n^2=1\) does converge. So when negative terms are allowed, the converse of one of the above results may not hold!

In other words, \(a_n^2\rightarrow A^2\) does not imply that \(a_n\rightarrow A\) if \(a_n<0\) is allowed. \(\square\)

Corollary 2.2.4 Suppose that \(\{a_n\}\) converges to \(A\), then \(\{ca_n\}\) converges to \(cA\).

show/hide proof

Proof. Note that this is a direct result from Theorem 2.2.1(b) by letting \(b_n=c\) a constant. Here is a proof though, and is essentially a proof of Theorem 2.2.1(b) for the special case that one of the sequences is constant.

If \(c=0\), then \(ca_n=0\) for all \(n\) and hence converges to \(cA=0\). Let \(c\neq0\) so that \(|c|>0\). Then for any \(\epsilon>0\) we have that \(\frac\epsilon{|c|}>0\). Since \(a_n\rightarrow A\), we can choose \(n^*\in\mathbb N\) such that for any \(n\geq n^*\) we have \(|a_n-A| < \frac\epsilon{|c|}\). Thus \(|ca_n-cA| =|c|\cdot|a_n-A|<|c|\cdot \frac\epsilon{|c|}=\epsilon\). \(\blacksquare\)

Example

\(a_n\rightarrow 2\) and \(b_n\rightarrow3\) then by Theorem 2.2.1(a) and Corollary 2.2.4 we know that \(a_n+5b_n\rightarrow17\). \(\square\)

Theorem 2.2.7 sequence \(a_n\) converges to \(0\), and a sequence \(b_n\) is bounded, then the sequence \(a_nb_n\) converges to \(0\).

show/hide proof

Proof. We know that there is a real number \(M\) such that \(|b_n|\leq M\) for all \(n\) since it is a bounded sequence.

Now let \(\epsilon>0\) and choose any \(n^*\) such that \(n\geq n^*\) implies that \(|a_n|<\frac\epsilon M\). Then we have that \(|a_nb_n|=|a_n|\cdot|b_n|\leq |a_n|\cdot M < \frac\epsilon M\cdot M=\epsilon\) for any \(n\geq n^*\). Thus \(a_nb_n\rightarrow0\). \(\blacksquare\)

Note that I have presented the above theorem out of order relative to the textbook. I rpesent the squeeze theorem next, which is an extremely powerful result.

Squeeze theorem

Theorem 2.2.6 (Sandwich or Squeeze Theorem) Suppose that \(\{a_n\}\), \(\{b_n\}\), and \(\{c_n\}\) are three sequences, and suppose that there exists \(n_1\) such that \(a_n\leq b_n\leq c_n\) for all \(n\geq n_1\). If \(a_n\) and \(c_n\) both converge to \(A\), then \(b_n\) also converges to \(A\).

show/hide proof

Proof. We have that \(|a_n-A|<\epsilon\) and \(|c_n-A|<\epsilon\) are eventually true. We can choose a cut-off index beyond which they are true simultaneously, in particular. This means that \(A-\epsilon<a_n\leq b_n\leq c_n< A+\epsilon\) for \(n\) sufficiently large. Therefor we can make \(|b_n-A|<\epsilon\) eventually true. We conclude that \(b_n\rightarrow A\). \(\blacksquare\)

Example:

Consider the sequence \(a_n=\frac{n^5}{(n+\sqrt n)^n}\). You might intuitively understand that this should converge to zero, but it does look quite complicated. Note that as long as \(n\geq6\) we have that \((n+\sqrt n)^n\geq(n+\sqrt n)^6>n^6\). Thus we can argue that \(\frac{n^5}{(n+\sqrt n)^n}<\frac{n^5}{n^6}=\frac1n\) is eventually true. In fact it is true for all \(n\geq6\). Also we have \(a_n>0\) for all \(n\). So we have \(0<a_n<\frac1n\) for all \(n\) sufficiently large. \(0\rightarrow0\) and \(\frac1n\rightarrow0\) so we have that \(a_n\rightarrow0\) by a squeeze. \(\square\)

Example:

Consider the sequence \(a_n=\frac{n-1}{n+\sqrt n}\). It is easy to see that this sequence converges to \(1\) by dividing the top and bottom by \(n\) and then using Theorem 2.2.1. But we’ll consider a squeeze theorem application just to see another example.

If we multiply the top and bottom by \(n-\sqrt n\) we get \(a_n=\frac{n-1}{n+\sqrt n}=\frac{n^2-n+\sqrt n-n\sqrt n}{n^2-n}\). Now as long as \(n>1\) this is valid. We also have that \[1-\frac1n+\frac{\sqrt n}{n^2}-\frac{n\sqrt n}{n^2}=\frac{n^2-n+\sqrt n-n\sqrt n}{n^2-n}=a_n=\frac{n-1}{n+\sqrt n}<\frac{n}{n+\sqrt n}<\frac{n}{n}=1\] Now we have that the far left side converges to \(1\), hence \(a_n\overset{n\rightarrow\infty}{\longrightarrow}1\) by a squeeze. \(\square\)

\[ \diamond \S \diamond \]