# 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.