(last updated: 4:09:51 PM, September 27, 2020)
The following results are not in your textbook. They are provided as a supplement. I would recommend familiarizing yourself with these results as they will be useful at times. For example, if you know that \((a_n)^r\) converges where \(r\) is some real number exponent, then under certain circumstances, we then know that \(a_n\) converges. This implication goes both ways under certain circumstances as well. You do not need to read the proofs of these results, but you might find them helpful or interesting. You text only gives a limited number of these results.
First we prove that \(a_n\rightarrow0\) is equivalent to \(a_n^r\rightarrow0\) for any \(r\in\mathbb R^+\).
Theorem Let \(a_n\) be a sequence with nonnegative terms and \(r\) a positive real number. Then \(a_n\rightarrow0\) is equivalent to \(a_n^r\rightarrow0\).
Proof. Assume that \(a_n\rightarrow 0\) (i.e. that \(A=0\)). Then for any \(\epsilon>0\) we can make \(|a_n|<\epsilon\) eventually true. In particular we can make \(|a_n|<\epsilon^{1/r}\) eventually true for \(n\) sufficiently large and any \(r\in\mathbb R\) with \(r>0\). We need \(r>0\) so that we avoid taking reciprocals of \(\epsilon\); if \(\epsilon\) is small, then its reciprocal becomes large. Now we have that \(|a_n|<\epsilon^{1/r}\) implies that \(|(a_n)^r|=|a_n|^r<(\epsilon^{1/r})^r=\epsilon\). Thus \(a_n^r\rightarrow0\). Since \(r\) is an arbitrary positive real number here, it can be any \(k\in\mathbb N\) or \(\frac1k\) or any \(q\in\mathbb Q\) as well.
Now assume that \(a_n^r\rightarrow 0\) for some \(r\in\mathbb R^+\). Then for any \(\epsilon>0\) we know we can make \(|a_n^r|<\epsilon\) eventually true. Thus we can make \(|a_n^r|<\epsilon^r\) eventually true. Note that \(\epsilon^r\) is still just an arbitrary positive number. Hence \(|a_n|=|a_n^r|^{1/r}<(\epsilon^r)^{1/r}=\epsilon\) and we have that \(a_n\rightarrow 0\).
Since \(a_n\geq0\), then \(a_n^r\geq0\) for all \(r>0\), thus we will always have \(A^r\geq0\). We have already completed the \(A=0\) case. So now we just need to work on the \(A>0\) case. \(\blacksquare\)
Proof. Assume \(A>0\) and \(a_n^k\rightarrow A^k\) for some \(k\in\mathbb N\). We will prove that \(a_n\rightarrow A\).
Since \(a_n^k\rightarrow A\), we can make \(0<\frac{A^k}{2}<a_n^k<\frac{3A^k}{2}\) eventually true. This implies that \(0<\left(\frac{1}{2}\right)^{1/k}A<a_n<\left(\frac{3}{2}\right)^{1/k}A\) is eventually true. Note that \[|a_n-A|=\frac{|a_n^k-A^k|}{\left| \displaystyle\sum_{l=0}^{k-1}a_n^{k-1-l}A^l\right|}\] We will find a positive lower bound on the denominator. Eventually, for large enough \(n\) we have that all terms in the series in the denominator are positive thus we can remove the absolute value bars and get \[\left| \sum_{l=0}^{k-1}a_n^{k-1-l}A^l\right|=\sum_{l=0}^{k-1}a_n^{k-1-l}A^l=S(k,n,A)\] Furthermore we have that, eventually, \[ 0<C_1=\sum_{l=0}^{k-1} \left(\left(\frac{1}{2}\right)^{1/k}A\right)^{k-1-l}A^l \geq \sum_{l=0}^{k-1}a_n^{k-1-l}A^l \leq \sum_{l=0}^{k-1} \left(\left(\frac{3}{2}\right)^{1/k}A\right)^{k-1-l}A^l=C_2\] Note that the left and right sides are just positive constants that do not dened on \(n\). We just require that \(n\) be sufficiently large.
Since \(a_n^k\rightarrow A^k\) we can make \(|a_n^k-A^k|<C_1\cdot \epsilon\) eventually true. Thus we have that \[|a_n-A|=\frac{|a_n^k-A^k|}{\left| \displaystyle\sum_{l=0}^{k-1}a_n^{k-1-l}A^l\right|}=|a_n^k-A^k|\cdot\frac{1}{\left| \displaystyle\sum_{l=0}^{k-1}a_n^{k-1-l}A^l\right|}<C_1 \epsilon\cdot \frac{1}{C_1}=\epsilon\] So we have show that \(a_n\rightarrow A\). And we have that (a) and (b) are equivalent when \(A>0\).
Now assume that \(b_n=\sqrt[k]{a_n}\rightarrow \sqrt[k]{A}=B\). Then we know that \(a_n=b_n^k\rightarrow B^k=A\). Thus we have that (a) and (c) are equivalent. \(\blacksquare\)
Lemma Let \(\epsilon>0\) and \(n\in\mathbb N\). Then \((1+\epsilon)^n\geq 1+n\epsilon\).
Proof. Just use the binomial expansion and note that all terms are positive: \((1+\epsilon)^n=1+n\epsilon+{n\choose2}\epsilon^2+{n\choose3}\epsilon^3+\cdots +{n\choose{n-2}}\epsilon^{n-2}+n\epsilon^{n-1}+\epsilon^{n}\geq1+n\epsilon\). \(\blacksquare\)
Theorem Let \(a>0\). Then \(a^{1/n}\rightarrow 1\).
Proof. Assume that \(a>1\) and that \(a=1+\delta\) for some \(\delta>0\). We consider the sequence \(a_n=a^{1/n}\). Then \(1<a^{\frac1{n+1}}<a^{\frac1{n}}<a\) for all \(n\in\mathbb N\). So \(a^{1/n}\) is a strictly decreasing sequence that is bounded from below by \(1\). If it converges, it cannot converge to any number below 1.
We show that \(|a^{1/n}-1|=a^{1/n}-1\) converges to zero. We also have that \((1+\epsilon)^n\geq 1+n\epsilon\) for any \(\epsilon>0\) and \(n\in\mathbb N\) (Lemma given above). Now choose any \(n^*\) such that \(\frac \delta {n^*}<\epsilon\) (this is possible by the archimedean property). Then if \(n\geq n^*\) we have that \(1+\delta<1+n^*\epsilon\leq 1+n\epsilon\leq (1+\epsilon)^n\). This implies that eventually \(a_n=(1+\delta)^{1/n}<1+\epsilon\) is true. This proves that \(a^{1/n}\rightarrow1\).
Now if \(0<a<1\) we have that \(\frac1a>1\) and thus that \(\left(\frac1a\right)^{1/n}\rightarrow1\) by the above argument. We also know that \(\lim_{n\rightarrow\infty}\left(\frac1a\right)^{1/n}=\frac1{\lim_{n\rightarrow\infty}a^{1/n}}=\frac11\).
If \(a=1\) then \(a^{1/n}=1\) for all \(n\). This completes the proof. \(\blacksquare\)
Proof. Assume that \(a_n\rightarrow A\) and \(q\in\mathbb Q\). Let \(q=\frac km\) for \(k,m\in\mathbb N\). Then \(b_n=(a_n)^k\rightarrow A^k=B\) and \(a_n^q=(b_n^{1/m})\rightarrow B^{1/m}=A^q\). Thus (a) implies (b).
Now assume that \(a_n^q\rightarrow A^q\) for some positive rational number \(q\). Let \(q=\frac km\) where \(k,m\in\mathbb N\). Then \(a_n^q=(a_n^k)^{1/m}\). We know that \(a_n^k=(a_n^q)^m\rightarrow (A^q)^m=A^k\), which in turn implies that \(a_n\rightarrow A\). Thus (b) implies (a), and we conclude the statements are equivalent. \(\blacksquare\)
Proof. If \(r\) is rational, then we have equivalence, so we are concerned only with irrational \(r\).
Note that an irrational number satisfies \(r=\sup\{q\in\mathbb Q \mid 0<q<r\}\) requiring that \(r>0\) of course. This means that for any irrational number \(r>0\) and any \(\epsilon\in\mathbb R\) satisfying \(0<\epsilon<r\) we can always find a rational number \(q\) such that \(0<r-\epsilon<q\) by the approximation property of suprema (Theorem 1.7.7). We will explore how \(x^q\) is close to \(x^r\),
Assume that \(a_n\rightarrow A>0\). Thus we know \(a_n\) is bounded by some \(M\). We also know that \(a_n^{1/k}\rightarrow 1\) as \(k\rightarrow\infty\). So we can choose a \(q\) that is so close to \(r\) but slightly smaller, \(0<q<r\), to make \(a_n^{r-q}\) and \(A^{r-q}\) as close to \(1\) as we like. In particular we can choose a \(q\) such that \(|a_n^{r-q}-1|<\frac\epsilon{3(1+M)^r}\) and \(|A^{r-q}-1|<\frac\epsilon{3(1+A)^r}\). We wish to bound \(a_n^q\), but we do not know if \(a_n\) is less than or greater than one or the same for \(r\) as well, so we use \(1+M\) as the bound to get \(a_n\leq M\) implies that \(|a_n|^q \leq (1+M)^{1+r}\) and \(A^q \leq (1+A)^{1+r}\). Now choose an \(n^*\) such that for any \(n\geq n^*\) we have \(|a_n^q - A^q|<\frac\epsilon3\).
Then we have that \[ \begin{aligned} |a_n^r-A^r| &=|a_n^qa_n^{r-q}-A^qA^{r-q}|\\ &=|a_n^qa_n^{r-q} -a_n^q+a_n^q -A^q +A^q -A^qA^{r-q}|\\ &=|a_n^q(a_n^{r-q} - 1) + (a_n^q - A^q) + A^q(1 - A^{r-q})|\\ &\leq|a_n^q|\cdot |a_n^{r-q} - 1| + |a_n^q - A^q| + |A^q|\cdot|1 - A^{r-q}|\\ &< |a_n^q|\cdot \frac\epsilon{3(1+M)^{r+1}} + \frac\epsilon3 + |A^q|\cdot\frac\epsilon{3(1+A)^{r+1}}\\ &\leq (1+M)^{r+1}\cdot \frac\epsilon{3(1+M)^{r+1}} + \frac\epsilon3 + (1+A)^{r+1}\cdot\frac\epsilon{3(1+A)^{r+1}}\\ &=\epsilon \end{aligned} \] Thus \(a_n^r\rightarrow A^r\).
Now to prove the other direction, assume that \(a_n^r\rightarrow A^r\). Then we know that \(r>0\) so that \(\frac1r>0\) also. We pply the previous result with \(\frac1r\) in place of \(r\) to get \(a_n=(a_n^r)^{1/r}\rightarrow (A^r)^{1/r}=A\). \(\blacksquare\)
Corollary If \(a_n>0\) for all \(n\), \(A>0\), and \(\alpha\in\mathbb R\) is any nonzero real number, then \(a_n^\alpha\rightarrow A^\alpha\) is equivalent to \(a_n\rightarrow A\).
No proof will be given for the above corollary, but it follows form similar arguments already given. If \(\alpha>0\), then it is already shown to be equivalent. If \(\alpha<0\), then consider the reciprocal sequence \(\frac1{a_n}\). Note that we require a strictly positive sequence here that converges to a strictly positive limit.
We have not explored the idea of continuity of functions yet, but these results essentially tell us that power functions, those of the form \(f(x)=x^\alpha\) for any \(\alpha\neq0\) are continuous functions. That is, that as \(x\rightarrow A\) we have that \(x^\alpha\rightarrow A^\alpha\). Of course, we will define this concept rigorously later on.
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