Sequences

Motivation

  • At the end of Strategy for integration, we have mentioned that there are elementary functions, such as \(f(x)=e^{-x^2}\), the antiderivative of which is not an elemetary function.
    • On the other hand, we'll see that there's a larger notion of functions, that of convergent power series, the antiderivative of which is going to be another convergent power series.
    • We'll see that although the antiderivative might not have a closed formula, that is a description as an elementary function, you can readily approximate its values using the power series description.
    • To introduce power series, and discuss their convergence, we first need to introduce sequences and their convergence.
    • This is a rather large section, but most of it is very similar to convergence of functions. Since we are on a tight schedule, we won't be able to cover all of it. Please read the textbook too.

Definition and notation

  • A sequence is a list of numbers written in a definite order: \[ a_1,a_2,\dotsc,a_n,\dotsc \]
    • The number \(a_1\) is the first term, the number \(a_2\) is the second term, and in general \(a_n\) is the \(n\)-th term.
    • The sequence \(\{a_1,a_2,a_3,\dotsc\}\) is also denoted by \[ \{a_n\}\text{ or }\{a_n\}_{n=1}^\infty. \]
  • Examples. Some sequences can be defined by giving a formula for the \(n\)-th term.
    • \(\left\{\frac{2n}{n+2}\right\}\), \(a_n=\frac{2n}{n+2}\), \(\left\{\frac23,1,\frac65,\dotsc,\frac{2n}{n+2},\dotsc\right\}\)
    • \(\left\{\frac{(-1)^n(n-2)}{2^n}\right\}\), \(a_n=\frac{(-1)^n(n-2)}{2^n}\), \(\left\{\frac12,0,-\frac18,\frac18,-\frac{3}{32},\dotsc,\frac{(-1)^n(n-2)}{2^n},\dotsc\right\}\)
    • \(\left\{\sqrt{2n+1}\right\}\), \(a_n=\sqrt{2n+1}\), \(\left\{\sqrt3,\sqrt5,\sqrt7,\dotsc\sqrt{2n+1},\dotsc\right\}\)
    • \(\left\{\sin\frac{n\pi}{4}\right\}\), \(a_n=\sin\frac{n\pi}{4}\), \(\left\{2^{-1/2},1,2^{-1/2},0,-2^{-1/2},\dotsc,\sin\frac{n\pi}{4},\dotsc\right\}\)

Examples.

  • Example. Let's find a formula for the sequence \[ \left\{\frac23,-\frac46,\frac89,-\frac{16}{12},\frac{32}{15}\right\}, \] assuming that the pattern continues.
    • Since the nominators are \(2,4,8,16,32\), in general the \(n\)-th nominator is \(2^n\).
    • Since the denominators are \(3,6,9,12,15\), in general the \(n\)-th denominator is \(3n\).
    • Since the signs are \(1,-1,1,-1,1\), in general the \(n\)-th sign is \((-1)^{n+1}\).
    • All in all, the sequence is \(a_n=-\frac{2^n}{3n}\).
  • There are sequences which don't have a simple formula.
    • The sequence \(\{t_n\}\), where \(t_n\) is the surface temperature of Lake Ontario at Toronto Harbourfront on January 1 of the year \(1900+n\) at noon.
    • The sequence \(\{p_n\}\), where \(p_n\) is the digit in the \(n\)-th decimal place of the number \(\pi\).
    • The Fibonacci sequence \(\{f_n\}\). This is probably the most famous recursive sequence. Recursive sequences also have formulas, but these formulas include lower index terms. In this particular case, we have \[ f_1=1,\,f_2=1\,\text{ and }f_n=f_{n-2}+f_{n-1}\text{ for }n>2. \] Therefore, the first few terms are \[ \{1,1,2,3,5,8,13,21,\dotsc\} \]

Convergence of sequences

  • Consider again the sequence \(a_n=\frac{2n}{n+2}\).
    • There's 2 plots you can assign to this.
    • First, there is a 1-dimensional plot, where you put on the real line the numbers \(a_n\). Click here for the plot. You can see that the points approach the number 2.
    • Second, there is a 2-dimensional plot of the points \((n,a_n)\). Click here for the plot. You can see that the points approach the line \(y=2\).
    • Consider \[ a_n=\frac{2n}{n+2}=\frac{2n+4-4}{n+2}=2-\frac{4}{n+2}. \] Note that \(\frac{4}{n+2}\) can be made as small as we like by taking \(n\) sufficiently large.
    • This in turn shows that the difference \(|a_n-2|\) can be made as small as we like by taking \(n\) sufficiently large. We indicate this by writing \[ \lim_{n\to\infty}a_n=2. \]

Convergence of sequences

  • Definition (imprecise version). A sequence \(\{a_n\}\) has the limit L, and we write \[ \lim_{n\to\infty}a_n=L\text{ or }a_n\to L\text{ as }n\to\infty, \] if we can make the terms \(a_n\) as close to \(L\) as we like by taking \(n\) sufficiently large. If \(\lim_{n\to\infty}a_n\) exists, we say the the sequence converges (or is convergent). Otherwise, we say that the sequence diverges (or is divergent).
  • Definition (precise version). A sequence \(\{a_n\}\) has the limit L, and we write \[ \lim_{n\to\infty}a_n=L\text{ or }a_n\to L\text{ as }n\to\infty, \] if for every \(\varepsilon>0\), there is a corresponding integer \(N\) such that \[ \text{if }n>N\text{ then }|a_n-L|<\varepsilon. \]
    • Usually we can prove convergence by other means. The precise version is more useful for proving that a given sequence is divergent.

Convergence of sequences

  • As an example, let's show \(\lim_{n\to\infty}\frac{2n}{n+2}=2\) using the precise definition.
    • Let's pick some arbitrary \(\varepsilon>0\). We want to find \(N>0\) so that if \(n>N\), then \[ |a_n-2|=\left|\frac{4}{n+2}\right|<\varepsilon. \]
    • Since its argument is positive, we can get rid of the absolute value sign. Then we can transform the inequality to \[ \frac{4}{\varepsilon}-2<n. \]
    • You can see that if \(n>\left\lceil\frac{4}{\varepsilon}\right\rceil\), then the inequality holds. Therefore, we can make \(N=\left\lceil\frac{4}{\varepsilon}\right\rceil\).
    • For example, if \(\varepsilon=\frac{1}{10}\), then we can take \(N=40\). You can see in this plot that if \(n>40\), then \(|a_n-2|<\frac{1}{10}\).

Convergence of sequences

  • Consider \(a_n=\sin\frac{n\pi}{2}\).
    • Based on its plot, the sequence is divergent.
    • A precise proof is as follows. We have \[ a_n=\begin{cases} 1 & n=4k+1,\\ 0 & n=4k+2\text{ or } n=4k,\\ -1 & n=4k+3. \end{cases} \] Note that we can find \(n\) arbitrarily large so that \(a_n=1,0,\text{ or }-1\).
    • Now let \(\varepsilon=\frac12\). If we had \(\lim_{n\to\infty}a_n=L\), then there must be an \(N\) so that if \(n>N\), then \(|a_n-L|<\frac12\). But this would require \(|1-L|<\frac12\) and \(|-1-L|<\frac12\) at the same time, which is impossible.
  • Exercises. 11.1: 39, 53, 54.

Sequence limit laws

  • Theorem. Let \(\{a_n\}\) be a sequence and \(f(x)\) be a function. If \(f(n)=a_n\) and \(\lim_{x\to\infty}f(x)=L\), then \(\lim_{n\to\infty}a_n=L\).
    • Since we already know about convergence of functions, this theorem is one of the main tools in proving that a sequence is convergent.
    • For example, if \(f(x)=\frac{2x}{x+2}\), then \(f(n)=\frac{2n}{n+2}\). Therefore, since for example l'Hospital's rule shows that \(\lim_{x\to\infty}\frac{2x}{x+2}=2\), we get \(\lim_{n\to\infty}\frac{2n}{n+2}=\infty\).
    • Exercises. 11.1: 23, 27, 29, 33, 43, 45, 47.
  • Theorem. Let \(\{a_n\}\) be a sequence. Suppose that \(\lim_{n\to\infty}|a_n|=0\). Then \(\lim_{n\to\infty}a_n=0\).
    • Note that why this is important is because if \(a_n=(-1)^nf(n)\) for some function \(f(x)\), then we can't have \(a_n=g(n)\) for any function \(g(x)\), since noninteger powers of \(-1\) are not real numbers.
    • Consider \(a_n=\frac{(-1)^n(n-2)}{2^n}\). Since \(\lim_{n\to\infty}|a_n|=\lim_{n\to\infty}\frac{n-2}{2^n}=0\), we have \(\lim_{n\to\infty}a_n=0\).
    • Exercises. 11.1: 35, 36.

Sequence limit laws

  • There are limit laws for sequences corresponding to limit laws for functions.
    • You can see the full list in the textbook.
    • For example, if \(\{a_n\}\) and \(\{b_n\}\) are convergent sequences, then we have \(\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n\).
    • Example. Since we have \(\lim_{n\to\infty}(-1)^n\frac{2^n}{3^n}=0\), we get \(\lim_{n\to\infty}\frac{3^n+(-2)^n}{3^n}=1\).
    • Warning. It's very important that these limit laws only work if the component sequences. For example, although \(\lim_{n\to\infty}(n-n)=0\), the expression \(\lim_{n\to\infty}n-\lim_{n\to\infty}n\) is divergent.
    • Exercises. 11.1: 20, 21, 19.
  • Squeeze theorem for sequences. Let \(\{a_n\}\), \(\{b_n\}\) and \(\{c_n\}\) be sequences, and \(n_0\) be a positive integer. Suppose that \(a_n\le b_n\le c_n\) for \(n\ge n_0\), and that \(\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n=L\). Then we have \(\lim_{n\to\infty}b_n=L\).
    • Example. Recall that \(n!=1\cdot2\cdot\dotsb\cdot n\). Consider the sequence \(a_n=\frac{n!}{n^n}\).
    • Note that we have \[ a_n=\frac{1\cdot2\cdot\dotsc\cdot n}{n\cdot n\cdot\dotsb\cdot n}=\frac{1}{n}\frac{2\cdot3\cdot\dotsb\cdot n}{n\cdot n\cdot\dotsb\cdot n}. \]
    • Since we have \(0\le\frac{2\cdot3\cdot\dotsb\cdot n}{n\cdot n\cdot\dotsb\cdot n}\le\frac{1}{n}\), we get \(0\le a_n\le\frac{1}{n}\).
    • Since we have \(\lim_{n\to\infty}\frac{1}{n}=0\), we get \(\lim_{n\to\infty}a_n=0\).
    • Exercises. 11.1: 55, 56.

Monotonic sequences

  • Definition. Let \(\{a_n\}\) be a sequence.
    • If we have \(a_n\le a_{n+1}\) for all \(n\ge1\), we say that the sequence is increasing. If the inequalities are strict, then we say that the sequence is strictly increasing.
    • If we have \(a_n\ge a_{n+1}\) for all \(n\ge1\), we say that the sequence is decreasing. If the inequalities are strict, then we say that the sequence is strictly decreasing.
    • If the sequence is increasing or decreasing, we say that it is monotonic. If it is strictly increasing or decreasing, we say it is strictly monotonic.
    • Example. We could see on its graph that the sequence \(a_n=\frac{2n}{n+2}\) is strictly increasing. Let us see how we can prove this.
    • We need \[ a_n=\frac{2n}{n+2}<\frac{2(n+1)}{n+1+2}=a_{n+1}. \]
    • Multiplying up by the denominators, we get \[ 2n(n+3)=2n^2+6n<2n^2+6n+4=2(n+1)(n+2). \]

Bounded sequences

  • Definition. Let \(\{a_n\}\) be a sequence.
    • If there is a number \(M\) such that \(a_n\le M\) for all \(n\ge1\), then we say that the sequence is bounded above.
    • If there is a number \(m\) such that \(m\le a_n\) for all \(n\ge1\), then we say that the sequence is bounded below.
    • If the sequence is bounded above and bounded below, then we say that it is bounded.
    • Example. We could see on its graph that the sequence \(a_n=\frac{2n}{n+2}\) is bounded. Let's prove this.
    • We have \(0<a_n\), which proves that it's bounded below.
    • We saw on the graph that \(a_n<2\). Let's prove this. We need \[ \frac{2n}{n+2}<2. \]
    • Multiplying by \((n+2)\), we get \[ 2n<2(n+2)=2n+4. \]

The Monotonic sequence theorem

  • Monotonic sequence theorem. Every bounded, monotonic sequence is convergent.
    • So this is also a way to prove that \(a_n=\frac{2n}{n+2}\) is convergent.
    • This can also be used to show that some recursive sequences are convergent.
  • For example, consider the sequence \(a_n\) defined by \[ a_1=1,\,a_{n+1}=2+\frac{a_n}{2}. \]
  • First, let's check that \(a_n\) is increasing. We can use induction for this.
    • \(n=1\): We have \(a_1=1\le\frac52=a_2\).
    • Induction step. We assume \(a_{n-1}\le a_n\), and we need \[ a_n=2+\frac{a_{n-1}}{2}\le 2+\frac{a_n}{2}=a_{n+1}, \] which you can see by subtracting 2, and multiplying by 2.
  • Since the sequence is increasing, we immediately get \(a_n\ge a_1=1\), which shows that it's bounded below.

The Monotonic sequence theorem

  • Now we want to show that the sequence is bounded above. Since we can choose \(M\) arbitrarily large, let's show that \(a_n\le100\) for all \(n\ge1\).
    • We'll apply induction again. We immediately get \(a_1=1\le100\).
    • Suppose that \(a_n\le100\). The we have \[ a_{n+1}=2+\frac{a_n}{2}\le2+50\le100. \]
  • Then the Monotonic sequence theorem shows that \(\lim_{n\to\infty}a_n=L\). Using the limit laws, we can find \(L\): \[ L=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}(2+\frac{a_n}{2})=2+\frac{L}{2} \] gives \(L=4\).
  • Exercises. 11.1: 81, 79, 83, 93.