W07 - Notes

Sequences

Videos

Review Videos

Videos, Math Dr. Bob:

• Infinite sequences: Definition; Squeeze Theorem
  • Extra: Infinite sequences: Various examples, arithmetic and geometric
  • Extra: Infinite sequences: Recursive sequences (like Fibonacci)

Link to original

01 Theory

Theory 1

A sequence is a rule that defines a term for each natural number $n\in \mathbb{N}$:

$$a_0,a_1,a_2,a_3,a_4,\dots$$

So a sequence is a function from $\mathbb{N}$ to $ℝ$.

Geometric sequence

A sequence is called geometric if the ratio of consecutive terms is some constant $r$, independent of $n$:

$$\frac{a_{n+1}}{a_n}=r\text{for every}n$$

The defining relation of a geometric sequence is equivalent to $a_{n+1}=a_n\cdot r$.

By plugging $a_1=a_0\cdot r$ into $a_2=a_1\cdot r$, we have $a_2=(a_0\cdot r)\cdot r=a_0\cdot r^2$. This plugging can be repeated $n$-times to get a formula for the $n^{\text{th}}$ term:

$$a_n=a_{n-1}\cdot r=a_{n-2}\cdot r^2=a_{n-3}\cdot r^3=\cdots =a_1\cdot r^{n-1}=a_0\cdot r^n$$

Therefore $a_n=a_0\cdot r^n$, and we have a formula for the general term of the sequence (the term with index $n$).

Starting point of a sequence

Note that sometimes the index (variable) of a sequence starts somewhere other than $0$. Most common is $1$ but any other starting point is allowed, even negative numbers.

Sometimes $c$ is used instead of $a_0$ in the formula for the general term of a sequence, thus $a_n=c{r}^n$. The ‘$c$’ notation is useful when the sequence starts from $n\ne 0$.

Extra - Fibonacci sequence

The Fibonacci sequence goes like this:

$$0,1,1,2,3,5,8,13,21,34,55,89,144,\dots$$

The pattern is:

$$F_n=F_{n-1}+F_{n-2}$$

This formula is a recursion relation, which means that terms are defined using the values of prior terms.

The Fibonacci sequence is perhaps the most famous sequence of all time. It is related to the Golden Ratio and the Golden Spiral:

Link to original

02 Illustration

Example: Geometric sequence - revealing the format

Geometric sequence: revealing the format

Find $a_0$ and $r$ and $a_n$ (written in the geometric sequence format) for the following geometric sequences:

(a) $a_n={(-{\displaystyle \frac{1}{2}})}^n$ $$ (b) $b_n=-3\left({\displaystyle \frac{2^{n+1}}{5^n}}\right)$ $$
(c) $c_n=e^{5-7n}$

Solution

(a)

Plug in $n=0$ to obtain $a_0=1$. Notice that $a_{n+1}/a_n=-1/2$ and so therefore $r=-1/2$. Then the ‘general term’ is $a_n=a_0\cdot r^n=1\cdot {(-1/2)}^n$.

---

(b)

Rewrite the fraction:

$$\frac{2^{n+1}}{5^n}\gg \gg 2\cdot {\left(\frac{2}{5}\right)}^n$$

Plug that in and observe $b_n=-6\cdot {(2/5)}^n$. From this format we can read off $b_0=-6$ and $r=2/5$.

---

(c)

Rewrite:

$$c_n\gg \gg e^5\cdot e^{-7n}\gg \gg e^5\cdot {\left(e^{-7}\right)}^n$$

From this format we can read off $c_0=e^5$ and $r=e^{-7}$.

Link to original

Series

Videos

Review Videos

Videos, Math Dr. Bob:

• Infinite series: Definitions, basic examples
• Geometric series and SDT: Geometric series, Simple Divergence Test (aka “Limit Test”)
• Infinite series: Various examples
  • Extra: Infinite series convergence: Telescoping series

Link to original

03 Theory

Theory 1

A series is an infinite sum that is created by successive additions without end. The terms are not added up “all at once” but rather they are added up “as $n$ increases” or “as $n\to \infty$.”

$$a_0+a_1+a_2+a_3+\dots =\sum _{n=0}^{\infty }{a}_n$$

Three of the most famous series are the Leibniz series and the geometric series:

$$\begin{array}{cccc}\text{Leibniz series:}& 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots +\frac{{(-1)}^n}{2n+1}+\cdots & =& \frac{\pi }{4}\\ & & & \\ \text{Geometric series:}& 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +{\left(\frac{1}{2}\right)}^n+\cdots & =& 2\end{array}$$

---

Partial sum sequence of a series

The partial sum sequence of a series is the sequence whose terms are the sums up to the given index:

$$S_N=a_0+a_1+\cdots +a_N=\sum _{n=0}^N{a}_n$$

These $S_N$ terms themselves form a sequence:

$$S_0,S_1,S_2,S_3,\dots$$

Link to original

04 Illustration

Example: Geometric series - total sum and partial sums

Geometric series - total sum and partial sums

The geometric series total sum $S$ can be calculated using a “shift technique” as follows: (1) Compare $S$ and $rS$:

$$\begin{array}{cc}S=& a_0+a_{0}r+a_0{r}^2+a_0{r}^3+\cdots \\ {\gg \gg }^{\times r}rS=& a_{0}r+a_0{r}^2+a_0{r}^3+a_0{r}^4+\cdots \end{array}$$

(2) Subtract second line from first line, many cancellations:

$$\begin{array}{cc}S& =a_0+a_{0}r+a_0{r}^2+a_0{r}^3+\cdots \\ -(rS& =a_{0}r+a_0{r}^2+a_0{r}^3+a_0{r}^4+\cdots )\\ \text{—}\text{—}\text{—}& \text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\\ S-rS& =a_0\end{array}$$

(3) Solve to find $S$:

$$S=\frac{a_0}{1-r}$$

Assumes convergence!

Note: this calculation assumes that $S$ exists, i.e. that the series converges.

The geometric series partial sums can be calculated similarly, as follows:

(1) Compare $S$ and $rS$:

$$\begin{array}{cc}{S}_N=& a_0+a_{0}r+a_0{r}^2+\cdots +a_0{r}^N\\ {\gg \gg }^{\times r}r{S}_N=& a_{0}r+a_0{r}^2+\cdots +a_0{r}^N+a_0{r}^{N+1}\end{array}$$

(2) Subtract second line from first line, many cancellations:

$$\begin{array}{cc}{S}_N=& a_0+a_{0}r+a_0{r}^2+\cdots +a_0{r}^N\\ -(r{S}_N=& a_{0}r+a_0{r}^2+\cdots +a_0{r}^N+a_0{r}^{N+1})\\ \text{—}\text{—}\text{—}& \text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\text{—}\\ S_N-r{S}_N& =a_0-a_0{r}^{N+1}\end{array}$$

(3) Solve to find $S_N$:

$$\begin{array}{c}{S}_N=a_0\frac{1-r^{N+1}}{1-r}\\ \\ =\frac{a_0}{1-r}-\frac{a_0}{1-r}{r}^{N+1}=S-S{r}^{N+1}\end{array}$$

(4) The last formula is revealing in its own way. Here is what it means in terms of terms:

$$\begin{array}{c}{a}_0+a_{0}r+\cdots +a_0{r}^N=\\ \\ & \begin{array}{cc}{a}_0+a_{0}r+a_0{r}^2+\cdots & \\ & -(a_0{r}^{N+1}+a_0{r}^{N+2}+\cdots )\end{array}\end{array}$$Link to original

Convergence

Videos

Review Videos

Videos, Math Dr. Bob:

• Infinite sequences convergence: Squeeze; Monotone Bounded
• Infinite sequences convergence: Examples sequences: convergent, monotonic, bounded

Link to original

05 Theory

Theory 1

A sequence has a limit if its terms tend toward a specific number, or toward $\pm \infty$.

When this happens we can write “${\lim }_{n\to \infty }{a}_n=L$” with some number $L\in ℝ$ or $L=\pm \infty$. We can also write “$a_n\to L$ as $n\to \infty$”.

The sequence is said to converge if it has a finite limit $L\in ℝ$.

Some sequences don’t have a limit at all, like $a_n=\cos n$:

Or $a_n=e^n$:

These sequences diverge. In the second case, there is a limit $L=\infty$, so we say it diverges to $+\infty$.

A sequence may have a limit of $\pm \infty$ but is still said to diverge.

Extra - Convergence definition

The precise meaning of convergence is this. We have $a_n\to L$ as $n\to \infty$ if, given any proposed error $\epsilon >0$, it is possible to find $N$ such that for all $n>N$ we have $|a_n-L|<\epsilon$.

When $L=\infty$, convergence means that given any $B>0$, we can find $N$ such that for all $n>N$ we have $a_n>B$.

Similarly for $L=-\infty$.

---

If the general term $a_n$ is a continuous function of $n$, we can replace $n$ with the continuous variable $x$ and compute the continuous limit instead:

$$\underset{n\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }{a}_x$$

If $a_x$ would be a differentiable function, and we discover an indeterminate form, then we can apply L’Hopital’s Rule to find the limit value. For example, if the indeterminate form is $0\cdot \infty$, we can convert it to $\frac{\infty }{1/0}=\frac{\infty }{\infty }$ and apply L’Hopital.

Link to original

06 Illustration

Example - L’Hopital’s Rule for sequence limits

L’Hopital’s Rule for sequence limits

(a) What is the limit of $a_n={\displaystyle \frac{\ln n}{n}}$? (b) What is the limit of $b_n={\displaystyle \frac{{(\ln n)}^2}{n}}$? (c) What is the limit of $c_n=n(\sqrt{n^2+1}-\sqrt{n^2})$? (d) What is the limit of $d_n=n{e}^{-n}$?

Solution

(a)

Identify indeterminate form $\frac{\infty }{\infty }$. Change from $n$ to $x$ and apply L’Hopital:

$$\underset{x\to \infty }{\lim }\frac{\ln x}{x}{\gg \gg }^{\frac{d}{dx}}\underset{x\to \infty }{\lim }\frac{1/x}{1}=0$$

---

(b)

Identify indeterminate form $\frac{\infty }{\infty }$. Change from $n$ to $x$ and apply L’Hopital:

$$\underset{x\to \infty }{\lim }\frac{{(\ln x)}^2}{x}{\gg \gg }^{\frac{d}{dx}}\underset{x\to \infty }{\lim }\frac{2\ln x\cdot \frac{1}{x}}{1}=2\underset{x\to \infty }{\lim }\frac{\ln x}{x}\stackrel{\text{(by (a) result)}}{=}0$$

---

(c)

(1) Identify form $\infty \cdot 0$ and rewrite as $\frac{\infty }{\infty }$:

$$n(\sqrt{n^2+1}-\sqrt{n^2})\gg \gg \frac{\sqrt{n^2+1}-\sqrt{n^2}}{1/n}$$

(2) Change from $n$ to $x$ and apply L’Hopital:

$$\frac{\sqrt{x^2+1}-\sqrt{x^2}}{1/x}\gg \gg \frac{\frac{1}{2}{(x^2+1)}^{-1/2}(2x)-1}{-1/x^2}$$

(3) Simplify:

$$\gg \gg \frac{-2x^3}{\sqrt{x^2+1}}+\frac{1}{x^2}\gg \gg \frac{-2x^5+\sqrt{x^2+1}}{x^2\sqrt{x^2+1}}$$

(4) Consider the limit:

$$\frac{-2x^5+\sqrt{x^2+1}}{x^2\sqrt{x^2+1}}\stackrel{x\to \infty }{⟶}\frac{-2x^5+x}{x^3}⟶-\infty$$

---

(d)

(1) Identify form $\infty \cdot 0$ and rewrite as $\frac{\infty }{\infty }$:

$$n{e}^{-n}\gg \gg \frac{n}{e^n}$$

(2) Change $n$ to $x$ and apply L’Hopital:

$$\frac{x}{e^x}\gg \gg \frac{1}{e^n}\stackrel{x\to \infty }{⟶}0$$ Link to original

Example - Squeeze theorem

Squeeze theorem

Use the squeeze theorem to show that $\frac{4^n}{n!}\to 0$ as $n\to \infty$.

Solution

(1) We will squeeze the given general term above $0$ and below a sequence $b_n$ that we must devise:

$$0\le \frac{4^n}{n!}\le b_n$$

(2) We need $b_n$ to satisfy $b_n\to 0$ and $\frac{4^n}{n!}\le b_n$. Let us study $\frac{4^n}{n!}$.

$$\frac{4^n}{n!}=\frac{4\cdot 4\cdot \cdots \cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4}{n(n-1)\cdots 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$$

(3) Now for the trick. Collect factors in the middle bunch:

$$\frac{4^n}{n!}=\frac{4}{n}(\frac{4}{n-1}\cdot \frac{4}{n-2}\cdot \cdots \cdot \frac{4}{7}\cdot \frac{4}{6}\cdot \frac{4}{5})\frac{4\cdot 4\cdot 4\cdot 4}{4\cdot 3\cdot 2\cdot 1}$$

(4) Each factor in the middle bunch is $<1$ so the entire middle bunch is $<1$. Therefore:

$$\frac{4^n}{n!}<\frac{4}{n}\cdot \frac{4^4}{4!}=\frac{1024}{24n}$$

Now we can easily see that $1024/24n\to 0$ as $n\to \infty$, so we set $b_n=1024/24n$ and we are done.

Link to original

07 Theory

Theory 2

Monotone sequences

A sequence is called monotone increasing if $a_{n+1}\ge a_n$ for every $n$.

A sequence is called monotone decreasing if $a_{n+1}\le a_n$ for every $n$.

In this context, ‘monotone’ just means it preserves the increasing or decreasing modality for all terms.

Monotonicity Theorem

If a sequence is monotone increasing, and bounded above by $B$, then it must converge to some limit $L$, and $L\le B$.

If a sequence is monotone decreasing, and bounded below by $B$, then it must converge to some limit $L$, and $L\ge B$.

Terminology:

• Bounded above by $B$ means that $a_n\le B$ for every $n$
• Bounded below by $B$ means that $B\le a_n$ for every $n$

Notice!

The Monotonicity Theorem says that a limit $L$ exists, but it does not provide the limit value.

Link to original

08 Illustration

Example - Monotonicity Theorem

Monotonicity

Show that $a_n=\sqrt{n+1}-\sqrt{n}$ converges.

Solution

(1) Observe that $a_n>0$ for all $n$.

Because $n+1>n$, we know $\sqrt{n+1}>\sqrt{n}$.

Therefore $\sqrt{n+1}-\sqrt{n}>0$

---

(2) Show $a_x$ is decreasing:

Replace $n$ with $x$: $a_x=\sqrt{x+1}-\sqrt{x}$ considered as a differentiable function.

Take derivative to show decreasing:

$$\begin{array}{c}\frac{d}{dx}{a}_x=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x}}\\ \\ \gg \gg \frac{2(\sqrt{x}-\sqrt{x+1})}{4\sqrt{x}\sqrt{x+1}}\end{array}$$

Denominator is $>0$. Numerator is $<0$. So $\frac{d}{dx}{a}_x<0$ and $a_x$ is monotone decreasing.

Therefore $a_n$ is monotone decreasing as $n\to \infty$.

---

(3) Since $a_n>0$ it is bounded below by $0$. It is monotone decreasing. Conclude that it converges.

Link to original

09 Theory

Theory 3

Series convergence

We say that a series converges when its partial sum sequence converges:

$$\text{“}\sum _{n=0}^{\infty }{a}_n\text{converges}\text{”}\text{MEANS:}\text{“}{S}_N\text{converges as }N\to \infty \text{”}$$

Let us apply this to the geometric series. Recall our formula for the partial sums:

$$S_N=a_0\frac{1-r^{N+1}}{1-r}$$

Rewrite this formula:

$$\gg \gg S_N=\frac{a_0}{1-r}-\frac{a_0}{1-r}{r}^{N+1}$$

Now take the limit as $N\to \infty$:

$$\underset{N\to \infty }{\lim }{S}_N=\text{“}\frac{a_0}{1-r}-\frac{a_0}{1-r}{r}^{\infty +1}\text{”}=\frac{a_0}{1-r}$$

So we see that $S_N$ converges exactly when $|r|<1$. It converges to $\frac{a_0}{1-r}$.

(If $|r|=1$ then the denominator is $0$, and if $|r|>1$ then the factor $r^{\infty +1}$ does not converge.)

Furthermore, we have the limit value:

$$\sum _{n=0}^{\infty }{a}_n\gg \gg \underset{N\to \infty }{\lim }{S}_N\gg \gg \frac{a_0}{1-r}\gg \gg S$$

This result confirms the formula we derived for the total sum $S$ of a geometric series. This time we did not start by assuming $S$ exists, rather we proved that $S$ exists. (Provided that $|r|<1$.)

Extra - Looking at $S$ and $S_N$

Notice that we always have the rule:

$$\begin{array}{cc}{S}_N& =S-r^{N+1}S\\ & \\ S_N& =\frac{a_0}{1-r}-\frac{a_0}{1-r}{r}^{N+1}\end{array}$$

This rule can be viewed as coming from partitioning the full series into a finite part $S_N$ and the remaining infinite part:

$$$$

We can remove a factor $r^{N+1}$ from the infinite part:

$$S-S_N=r^{N+1}(a_0+a_{0}r+a_0{r}^2\dots )$$

The parenthetical expression is equal to $S$, so we have the formula $S_N=S-r^{N+1}S$ given above.

Link to original