(last updated: 3:41:39 PM, October 06, 2020)
You have now seen a variety of sequence theorems and are familiar with sequences converging to finite real numbers. Now we will look at some specific ways that sequences can diverge. In particular, squences that go off to plus or minus infinity.
Definition 2.4.1 A sequence an is said to be
Note that we define the same terms for decreasing sequences with inequalities appropriately flipped.
The statements in the following remark should be readily understood, but they are stated in order to call attention to them as different ways you can show that a sequence is increasing. Of course, you can work out similar statements for decreasing seuqences too.
Assume that an→A. We have already discussed that removing a finite number of terms does not affect the long term behavior. Thus an+k→A as n→∞ for any fixed constant k∈N as well. This an+k is identical to an except that the first k terms have been discarded. It is like a “shifted” version of an.
Firthermore we know that an→A∈R also implies that akn→Ak for fixed exponent k∈N and that a1kn→A1k. Of course we also can multiply by constants or add constants and we know how the limiting behavior is determined.
Example:
Assume an→A∈R with A≠0 and A>−1. Find the limit of bn=√an+1+an+2−1(an+5)2. We know that bn→√A+1+A−1A2.
Passing the limit through operations in a formula is related to the idea of that operation being continuous (which we will rigorously develop in the next chapter). For now you can do that with polynomials and ratios of polynomials (as long as you do not generate indeterminate forms or zeros in a denominator). You can actually do this with rational and irrational exponents now as well, since I have defined what that means in my supplementary notes on exponents. Again, you need to watch out for things like getting a negative underneath an even root, or flipping an inequality due to multiplication or division with a negative though.
You will see recursively defined sequences often: an+1=f(an) is a common type. For example, consider an+1=(an−2)2 with a1 unknown. If this sequence is to converge to a real number A, then we know that A=(A−2)2. Solving this will give A=1,4. This does not guarantee convergence be any means though!
Consider the case a1=0, then a2=4 and a3=4, so we can see that an=4 for all further n values. So for this choice of a1, the sequence does indeed converge to A=4.
Try a1=1, then an=1 for all n, so the sequence converges to A=1.
Try a1=5, then a2=9, a3=49, a4=472, etc. This sequence seems to grow without bound! This would need to be proved though. What we do know is that the sequence can only converge to A=1,4 and no other possibilities for convergence exist. It could diverge to ±∞ or even oscillate. We cannot ignore any possibility without careful proof.
The next theorem gives us a condition for knowing that convergence (or divergence) is certain.
Proof of (a). Let S={an∣n∈N} which is a set of real numbers and is bounded above by M. Recall that by Definition 1.7.6, Completeness of R, we have that sup. Let A=\sup S. Now by Theorem 1.7.7, the approximation property of suprema, we have that for any \epsilon>0 we can find an s\in S such that A-\epsilon<s. However, we know that s=a_m for some m\in\mathbb N. We also know that for any n>m we have that a_n\geq a_m since a_n is increasing. Now let n^*=m. Then for any n\geq n^* we have that a_n\geq a_{n^*}. Furthermore we have s=a_{n^*}, thus A-\epsilon<a_{n^*}. This gives A-a_{n^*}\leq A-a_{n}<\epsilon. Finally this tells us that |a_n-A|=A-a_n<\epsilon. Thus a_n\rightarrow A.
Proof of (b). We are given that a_n is unbounded which means that there is no real number M such that |a_n|\leq M for all n. In fact, since a_n is increasing, we know that a_1 is a lower bound. We also know that a_n is eventually positive since otherwise zero would be an upper bound (making the sequence bounded). So given any M>0 there exists an m\in\mathbb N such that a_m > M. Otherwise the sequence would be bounded below by a_1 and bounded above by M>0, and hence bounded. Now let n^*=m. Since the sequence is increasing, we know that M<a_{n^*}=a_m\leq a_n for all n\geq n^*. Hence a_n diverges to +\infty. \blacksquare
Remark (not in text) A sequence that is increasing and bounded form above converges. A sequence that is decreasing and bounded from below converges.
Here is the general process for monotone sequences. This is assuming the sequence is increasing. The same steps work for a decreasing sequence with inequalities appropriately reversed.
Show it is increasing using induction. Show that a_1\leq a_2, and then show that a_{n-1}\leq a_n implies that a_{n}\leq a_{n+1}.
Show it is bounded from above. Show that a_1\leq M for some M, and then show that a_n\leq M implies that a_{n+1}\leq M. Thus by induction the entire sequence is bounded above by M.
Since it is increasing and bounded from above we know it converges by the monotone convergence theorem. Assume a_n\rightarrow A and “take the limit” of the recursive formula and solve for A.
Example Consider the sequence defined recursively by a_1=0 and a_{n+1}=\sqrt{a_n+2}+3 for n\geq1. Prove convergence, and find the limit.
Proof.
Show it is increasing:
We have that a_2=\sqrt{2}+3>0=a_1. Now assume that a_n>a_{n-1}. Therefor a_{n+1}=\sqrt{a_n+2}+3>\sqrt{a_{n-1}+2}+3=a_n. We have show that a_n>a_{n-1} implies that a_{n+1}>a_n. Thus, by induction, a_{n+1}>a_n for all n. So we have that this sequence is in fact strictly increasing.
Show it is bounded from above:
We need to find a good bound. We have a_1=0 and a_2=\sqrt{2}+3<5. Let’s try a_n\leq 5. If a_n\leq5 then a_{n+1}=\sqrt{a_n+2}+3 \leq\sqrt7+3\approx 5.65, so that doesn’t work. Maybe we can make the square root go away. Try a_n\leq7 because 7+2=9 has a nice square root. and \sqrt9+3=6<7. We have that a_n\leq7 implies that a_{n+1}=\sqrt{a_n+2}+3 \leq\sqrt9+3=6 \leq 7. So we now have that a_n is bounded from above by 7 for all n.
Now since this sequence is increasing and bounded from above, we know it converges by the monotone convergence theorem.
Find the limit:
Since a_n\rightarrow A\in\mathbb R, we can take the limit of both sides of the recursive formula to get A=\sqrt{A+2}+3. Now solve for A: (A-3)^2=A+2 so A^2-6A+9=A+2 and A^2-7A+7=0. Thus A=\frac{7\pm\sqrt{49-4\cdot7}}{2}=\frac72\pm\frac{\sqrt{21}}{2}\approx1.208712,5.791288.
We know that a_2>3 and the sequence is increasing, so A\neq\frac72-\frac{\sqrt{21}}{2}\approx 1.2087124. Thus A=\frac72+\frac{\sqrt{21}}{2}.
So we have proved that a_n\rightarrow \frac72+\frac{\sqrt{21}}{2}. \square
Example Consider the sequence defined recursively by a_1\in(0,1) and a_{n+1}=1-\sqrt{1-a_n} for n\geq1. Consider all cases, prove convergence, and find the limit.
I present two different proofs. The first uses the monotone convergence theorem and not any \epsilon-based arguments. The second is a direct proof using an \epsilon-argument.
Proof 1. Note that x\in(0,1) implies that 1-x\in(0,1). Also recall that x\in(0,1) implies that 0<x<\sqrt x<1. Rearranged the recursive formula to see that 1-a_{n+1}=\sqrt{1-a_n}. We can reason that 1-a_n\in(0,1) for all n so that gives us that 1-a_n< \sqrt{1-a_n}=1-a_{n+1}. In this way we can see that 1-a_n is an increasing sequence, which tells us that a_n is a decreasing sequence. We can also see that \sqrt{1-a_n}\in(0,1) for all n, so it is a bounded sequence. Thus a_n converges by the monotone convergence theorem. Let a_n\rightarrow A.
Note that the recursive definition a_{n+1}=1-\sqrt{1-a_n} is a relationship between the sequence b_n=a_{n+1} for n\geq1 and a_n. The sequence b_n is identical to a_n except that it starts one term in “the future.” Since a_n\rightarrow A we thus know that b_n\rightarrow A. we discussed int eh past that discarding a finite number of terms of a sequence does not impact its convergence behavior. Taking the limit of both sides of the equation b_n=1-\sqrt{1-a_n} then gives A=1-\sqrt{1-A}. So the limit A must satisfy this equation! Solving for A gives 1-A=\sqrt{1-A}. Note that we already know that A\in[0,1]; since a_n\in(0,1) for all n it can’t possibly converge to a negative number some something beyond 1. Thus we get (1-A)^2=1-A \ \Rightarrow \ (1-A)^2-(1-A)=0 \ \Rightarrow \ (1-A)((1-A)-1)=0 \ \Rightarrow \ (1-A)(-A)=0** thus A=0,1 are the only possibilities. Note that you still need to be extremely careful even when solving simple algebra equations like this! We know that a_n<1 for all n and is decreasing, so it cannot possibly converge to 1. Hence a_n\rightarrow0. \square
Proof 2. Let b_n=1-a_n for all n. We will examine the sequence b_n instead.
Also note that b_n\in(0,1) for all n, and is an increasing sequence (by the same argument as in proof 1 that a_n is decreasing). Finally let c_n=\frac1{b_n}, and this c_n is a decreasing sequence. Note that c_n>1 for all n. So c_n is a decreasing sequence and is bounded below by 1 and hence converges by the monotone convergence theorem. Of course, we have that c_n=\frac{1}{1-a_n} and hence c_n\rightarrow\frac1{1-A} where a_n\rightarrow A. We will show that c_n\rightarrow1.
Let c_1=1+\delta>1. Note that c_1=\frac1{1-a_1} so that \delta=\frac1{1-a_1}-1. Now given that c_{n+1}=\sqrt{c_n}, we can show that c_n=(c_1)^\frac{1}{2^{n-1}}=(1+\delta)^\frac{1}{2^{n-1}}. Now we will argue that (1+\delta)^\frac{1}{2^{n-1}}\rightarrow1.
We want to show that (1+\delta)^\frac{1}{2^{n-1}}-1<\epsilon (note that no absolute value bars are necessary since the quantity on the right is positive). This is equivalent to 1+\delta<(1+\epsilon)^{2^{n-1}}. By the Bernoulli inequality (1+\epsilon)^{2^{n-1}}\geq 1+2^{n-1}\cdot \epsilon so we only need to get 1+\delta<1+2^{n-1}\cdot \epsilon which will occur if we can find an n so large that \frac{\delta}{\epsilon}<2^{n-1}. Normally, I would take a logarithm here, but I wish to avoid the use of logarithms since we have not formally introduced them.
Note thate 2^k=(1+1)^k=1+k+\cdots+k+1 by the binomial expansion. And of course, 1+k+\cdots+k+1>1+k for all k. So we have that 2^{n-1}>1+(n-1)=n for all n. So we just need to choose an n^*>\frac{\delta}{\epsilon} which will in turn make \frac{\delta}{\epsilon}<2^{n-1} for all n\geq n^*. This then implies that (1+\delta)^\frac{1}{2^{n-1}}-1<\epsilon for all n\geq n^*. We conclude that c_n=(1+\delta)^\frac{1}{2^{n-1}} converges to 1.
We have shown that c_n\rightarrow1 for c_n=\frac{1}{1-a_n}. We already know that a_n\rightarrow A by monotone convergence (see proof 1). Thus we have that 1=\frac1{1-A} showing us that A=0. \square
Theorem 2.4.8 (Archimedean Order Property of \mathbb R) Let x be any real number. Then there exists a positive integer n^* greater than x.
Proof. Let x\in\mathbb R be an arbitrary real number. Assume that a_n=n is bounded above by x. Then a_n\leq x for all n\in\mathbb N. Since a_n is a strictly increasing sequence and is bounded from above by x, we know it converges by the monotone convergence theorem. Assume a_n\rightarrow A. However, we know recursively that a_{n+1}=a_n+1, thus taking the limit of this recursive formula gives A=A+1. This is a contradiction, so a_n cannot be bounded above by x. \blacksquare
We already proved this (Theorem 1.7.8), but it is presented here again since it is an important property of the real number system.
Here is an example that is kind of strange, but I thought it was interesting, plus in the proof I use the Archimedean property.
Example Consider the sequence defined recursively by a_1\geq0 and a_{n+1}=(1+a_n)^{1+a_n} for n\geq1. Prove that this sequence diverges to infinity.
Proof. Clearly a_n\geq0 for all n because (1+a_n)^{1+a_n}\geq 1+a_n\geq1 when a_n\geq0. Also, a_{n+1}=(1+a_n)^{1+a_n}\geq 1+a_n>a_n for any a_n\geq0, thus this sequence is strictly increasing. Now we just need to show it is not bounded from above. Notice that a_n>1 for n\geq3 no matter what we choose for a_1. This means that a_{n+1}=(1+a_n)^{1+a_n}> 1+a_n for n sufficiently large. So, in summary, we know that there is an m\in\mathbb N such that a_m>1. This implies that a_{m+1}>1+a_m>1+1=2. Going one step further shows a_{m+2}>1+a_{m+1}>1+2=3. Using induction we can show that a_{m+k}>1+k for all k\in\mathbb N. For any M>0, there exists a natural number k such that M<1+k (Archimedean property), there for there is an index m+k such that a_{m+k}>1+k>M. we conclude that M cannot be an upper bound on the sequence a_n. Therefore since a_n is increasing and cannot be bounded from above, we know it diverges to +\infty by Theorem 2.4.4.
\diamond \S \diamond