Sequences and Series

1. Definition of a sequence and a series

What is a sequence, what is a series of a sequence? Explain.
The sequence $(a_n)$ is given by $a_n=2n-1\quad (n=1,2,3,...)$ the sequence $(b_n)$ is given by $b_1=1, b_{n+1}=b_n+2\quad (n=1,2,3,...)$
1. Which representation is called explicit, which recursive?
2. Determine for $(a_n)$ the first 4 terms.
3. Determine for $(b_n)$ the first 4 terms.
The sequence $(c_n)$ is given by $c_1=2, c_{n+1}=5c_n \quad (n=1,2,3,...)$
1. Determine the first $4$ terms. What is the name of this sequence?
2. Determine $s_3$ , where $(s_n)$ is the series of the sequence.
The sequence $(d_n)$ is given by $d_1=2, d_{n+1}=d_n+5 \quad (n=1,2,3,...)$
1. Determine the first $4$ terms. What is the name of this sequence?
2. Determine $s_4$ , where $(s_n)$ is the series of the sequence.

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Solution 1

A sequence is a list of numbers or terms: $(a_n)=a_1, a_2, a_3, ...$ Often we are interested in the sum of the first $n$ terms: $s_n=a_1+a_2+...+a_n$ where $n=1,2,3,...$ , thus $s_1=a_1, s_2=a_1+a_2, s_3=a_1+a_2+a_3$ and so on. We call this new sequence $(s_n)=s_1, s_2, s_3, ...$ the series of the sequence $(a_n)$
We have
1. $(a_n)$ is explicit, $(b_n)$ is recursive.
2. $a_1=2\cdot 1-1=1, a_2=2\cdot 2-1=3, a_3=2\cdot 3 -1=5, a_4=2\cdot 4-1=7, ...$ (the odd numbers)
3. $b_1=1, b_2=1+2=3, b_3=3+2=5, b_4=5+2=7, ...$ (the odd numbers)
We have
1. $c_1=2, c_2=5\cdot 2=10, c_3=5\cdot 10=50, c_4=5\cdot 50=250, ...$ (geometric sequence)
2. $s_3=2+10+50=62$
We have
1. $d_1=2, d_2=2+5=7, d_3=7+5=12, d_4=12+5=17, ...$ (arithmetic sequence)
2. $s_4=2+7+12+17=38$

2. Arithmetic and geometric sequences

Consider the two sequences

(a_n)=2, 2.5, 3, 3.5, ...

and

(b_n)=2, 2.5, 3.125, 3.90625, ...

Which of these sequences is arithmetic, which geometric? Why? What is the common difference, what is the common quotient?
Calculate $a_5$ and $b_5$ .
Determine the explicit form for both sequences. Explain why these formulas hold.
Calculate with the explicit formulas $a_{12}$ and $b_{12}$ .
For which $n$ is $a_n$ greater than $1000$ for the first time?
For which $n$ is $b_n$ greater than $1000$ for the first time?
Which sequence describes linear growth, which exponential growth? And what are the respective functional equations describing this growth?

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Solution 2

$(a_n)$ is an arithmetic sequence, because from one term to the next always the same number $d$ is added ( $d=0.5$ ). $d$ is called the common difference, because it is $d=a_2-a_1=a_3-a_2= ...$ $(b_n)$ is a geometric sequence, because from one term to the next always the same factor $d$ is multiplied ( $q=1.25$ ). $d$ is called the common quotient, since it is $q=\frac{a_2}{a_1}=\frac{a_3}{a_2}= ...$
$a_5=a_4+d=3.5+0.5=4$ , $b_5=q\cdot a_4 = 1.25\cdot 3.90625= 4.8828125$ .
We have $a_n=2+(n-1)\cdot 0.5$ (since from $a_1$ to $a_n$ the number $0.5$ is added exactly ( $n-1$ )-times to $2$ ) and $b_n=2\cdot 1.25^{n-1}$ (since from $b_1$ to $b_n$ the number $1.25$ is multiplied exactly ( $n-1$ )-times with $2$ ).
$a_{12}=2+11\cdot 0.5=7.5$ , $b_{12}=2\cdot 1.25^{11}=23.283...$
Find $n$ with $a_n=2+(n-1)\cdot 0.5=1000$ It follows $n=1997$ . So the answer is $n=1998$ .
Find $n$ with $b_n=2\cdot 1.25^{n-1}=1000$ So $1.25^{n-1}=500$ . Apply the logarithm to both sides. It follows $n=\frac{\ln(500)}{\ln(1.25)}+1=28.85$ So the answer is $n=29$ .
The arithmetic sequence $(a_n)$ describes linear growth. At "time" $x=1$ the "population size" is $y=2$ , at "time" $x=2$ the "population size" is $y=2.5$ . It follows $f(x)=2+0.5\cdot \frac{x-1}{1}=0.5x+1.5$ The geometric sequence $(b_n)$ describes exponential growth. At "time" $x=1$ the "population size" is $y=2$ , at "time" $x=2$ the "population size" is $y=2.5$ . It follows $f(x)=2\cdot 1.25^\frac{x-1}{1}=2\cdot 1.25^{x-1}$

3. Geometric and arithmetic series (finite sums)

What is the summation formula for summing the first $n$ terms $s_n=a_1+a_2+...+a_{n}$ of an arithmetic sequence $(a_n)$ ?
What is the summation formula for summing the first $n$ terms $s_n=b_1+b_2+...+b_n$ of a geometric sequence $(b_n)$ ?
Calculate $s_{10}$ for the sequence $(a_n)=2, 2.5, 3, 3.5, ...$ . For which $n$ is $s_n$ greater than $10000$ for the first time?
Calculate $s_{10}$ for the sequence $(b_n)=2, 2.5, 3.125, 3.90625, ...$ . For which $n$ is $s_n$ greater than $10000$ for the first time?
Calculate the sum $20+19.2+...+1.6$ (assuming that the numbers form an arithmetic sequence).
Calculate the sum $16+12+9+...+\frac{59049}{65536}$ (assuming that the numbers form a geometric sequence).

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Solution 3

The summation formula for an arithmetic sequence $(a_n)$ is $s_n=\frac{n(a_1+a_n)}{2}$ (half of $n$ times the of the first and last term).
The summation formula for a geometric sequence $(b_n)$ is $s_n=a_1\cdot\frac{1-q^n}{1-q}$ where $q$ is the common quotient of $(b_n)$ .
$a_{10}=2+9\cdot 0.5=6.5$ , also $s_{10}=\frac{10(2+6.5)}{2}=42.5$ Find $n$ with $s_n>10000$ : As we have $a_n=2+(n-1)\cdot 0.5$ , find $n$ with $s_n=\frac{n(2+2+(n-1)0.5)}{2}=\frac{n(3.5+0.5n)}{2}$ Solve the equation for $n$ $\frac{n(3.5+0.5n)}{2}=10000$ This is a quadratic equation. Using the midnight formula, we get $n_1=-203.5$ und $n_2=196.5$ . Thus the solution is $n=197$ .
With $q=1.25$ follows $s_{10}=2\cdot \frac{1-1.25^{10}}{1-1.25}=66.505...$ Find $n$ with $s_n>10000$ , that is, find $n$ with $s_n=2\cdot \frac{1-1.25^n}{1-1.25}=10000$ It follows $1.25^n = 1251$ Applying the logarithm on both sides, we get $n=\frac{\ln(1251)}{\ln(1.25)}=31.96$ Thus, the solution is $n=32$ .
First, find $n$ : We know that $d=19.2-20=-0.8$ , thus $a_n=20+(n-1)(-0.8)=1.6$ It follows $n=24$ . Now we have $\underbrace{20}_{a_1}+19.2+...+\underbrace{1.6}_{a_{24}}=s_{24}=\frac{24(20+1.6)}{2}=259.2$
First, find $n$ : We know that $q=\frac{12}{16}=0.75$ , thus $a_n=16\cdot 0.75^{n-1}=\frac{59049}{65536}$ and it follows $n=11$ (apply logarithm). Therefore we have $\underbrace{16}_{a_1}+12+9+...+\underbrace{\frac{59049}{65536}}_{a_{11}}=s_{11}=16\frac{1-0.75^{11}}{1-0.75}=61.29$

4. Limits of arithmetic and geometric sequences

What does it mean if a "sequence converges", and a "sequence diverges"? What is the limit of a sequence? Explain.
Give examples of convergent sequences with
1. limit $0$
2. limit $3$ .
Give examples of divergent sequences that
1. strive to $\infty$ ,
2. strive to $-\infty$ ,
3. alternate between $-1$ and $1$ .
4. alternate between increasing and alternating positive and negative numbers.
When does an arithmetic sequence converge?
When does a geometric sequence converge? If the geometric sequence converges, what is its limit?

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Solution 4

If we keep increasing $n$ and the terms $a_n$ tend towards a certain numerical value $a$ , we say that the sequence converges (more precisely: converges towards $a$ ). The numerical value $a$ is called the limit of the sequence, and we write $\lim_{n\rightarrow \infty} a_n = a$ or $a_n\rightarrow a\quad (n\rightarrow \infty)$ or a bit more casual $a_\infty=a$ If a sequence does not converge (for example, if the terms $a_n$ oscillate between two values, or tend towards $\infty$ or $-\infty$ ), we say that the sequence diverges.
A few examples:
1. $a_n=\frac{1}{n}$ converges to $0$ .
2. $b_n=3-\frac{1}{n}$ is convergent (converges to $3$ )
A few examples:
1. $a_n=1000-n^2$ is divergent (tends towards $\infty$ ).
2. $b_n=1-2n$ is divergent (tends to $-\infty$ )
3. $c_n=(-1)^n$ is divergent (alternates between $-1$ and $1$ )
4. $d_n=(-2)^n$ alternates between increasing positive and negative numbers.
Arithmetic sequences never converge, so they are always divergent. In fact, they always tend to $\infty$ or $-\infty$ . This is easily seen by inspecting the explicit form $a_n=a_1+(n-1)d$ . An except is the uninteresting case where the common difference $d=0$ . In this case, the sequence is convergent.
Geometric sequences converge only if $-1<q<1$ holds for the common quotient $q$ . If this is the case, then $a_n=a_1\cdot q^{n-1}$ converges to $0$ . For all other values of $q$ ( $q$ bigger or equal than $1$ or smaller or equal to $-1$ ) $a_n$ tends toward $\infty$ , towards $-\infty$ , or shows some kind of oscillatory behaviour (e.g. check out sequences $e_n$ and $f_n$ above).

5. Limits of arithmetic and geometric series (infinite sums)

Consider a sequence $(a_n)$ and its series $(s_n)$ . Give an interpretation of $s_\infty$ .
When is the infinite sum of an arithmetic sequence finite?
When is the infinite sum of a geometric sequence finite? How is this sum calculated?
Calculate the infinite sum $2+1.9+...$ , where the numbers form
1. an arithmetic sequence.
2. a geometric sequence.
Calculate the infinite sum where the numbers form a geometric sequence:
1. $2+2.1+...$
2. $2-1.9+-...$
3. $2-2.1+-...$

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Solution 5

We have $\begin{array}{lll} s_1 & = &a_1\\ s_2 & =& a_1+a_2\\ s_3 & =& a_1+a_2+a_3\\ ... & & ...\\ s_\infty & =& a_1+a_2+a_3+a_4+... \end{array}$ Thus, the limit $s_\infty$ of the series $(s_n)$ is the sum of the infinite many terms of the sequence $(a_n)$ . In short, $s_\infty$ is an infinite sum. If $(s_n)$ converges, the sum $a_1+a_2+...$ is finite. If $(s_n)$ diverges, the sum $a_1+a_2+...$ is either $\infty$ or $-\infty$ , or does not exist (for oscillatory behaviour).
Never, as the terms $a_n$ of an arithmetic sequence tend to $\infty$ or $-\infty$ . Thus, the sum of these terms can not be finite. The exception is the uninteresting case where all terms of the sequence $(a_n)$ are zero.
Only for $-1<q<1$ . Then we have $s_\infty = a_1\cdot\frac{1}{1-q}$ where $q$ is the common quotient of the sequence $(a_n)$ . This formula follows from the summation formula $s_n=a_1\cdot\frac{1-q^n}{1-q}$ Indeed, for $-1<q<1$ , the value $q^n$ converges towards $0$ for $n\rightarrow \infty$ , so $s_n\rightarrow a_1\cdot\frac{1-0}{1-q}$ Important: Obviously you can insert any $q$ -value into the formula $s_\infty = a_1\cdot\frac{1}{1-q}$ and you will get a value. But only if $-1<q<1$ will this value correspond to the infinite sum of the geometric sequence. See the example below.
We have:
1. $s_\infty=-\infty$
2. $q=\frac{1.9}{2}=0.95$ and as $-1<q<1$ it follows $s_\infty=2\cdot\frac{1}{1-0.95}=40$ .
We have:
1. $q=\frac{2.1}{2}=1.05$ and thus it is not $-1<q<1$ . It follows that the infinite sum is not finite (indeed it is $\infty$ ). Note that we can still use the formula $s_\infty=2\cdot\frac{1}{1-1.05}=-40$ . But the result is wrong and is not the infinite sum.
2. $q=\frac{-1.9}{2}=-0.95$ and from $-1<q<1$ follows $s_\infty=2\cdot\frac{1}{1+0.95}=1.025...$ .
3. $q=\frac{-2.1}{2}=-1.05$ and $-1<q<1$ does not hold. So the infinite sum is not finite, and also not $\infty$ or $-\infty$ . Indeed, the terms $s_n$ become larger and larger (in the positive as well as in the negative range). Again note that we can still apply the formula $s_\infty=2\cdot\frac{1}{1+1.05}=0.975$ . But this is not the infinite sum.

6. Geometric problems

The following are the first three figures of an infinite sequence of figures. Figure 1 is a square of side length 3. To get figure 2 three squares of side length $1$ are added, and so on. The limit figure $F_\infty$ consists of infinitely many squares.
1. determine the area (sum of the square areas) of $F_\infty$ .
2. in which figure is the height of the object $99.99\%$ the height of $F_\infty$ ?
Adding smaller and smaller equilateral triangles as shown in the figure below, we obtain a limit object $F_\infty$ consisting of infinitely many equilateral triangles. The first triangle has side length $1$ . After each iteration, the side length is halved.
1. Which equilateral triangle has for the first time a side length smaller than $0.0001$ ?
2. Determine the length $L_\infty$ of the fat spiral $a_1, a_2, ...$ .
3. Let $L_n$ be the sum of the first $n$ side lengths. For which $n$ is $L_n$ for the first time greater than $1.9999$ ?
4. Determine the area $A_\infty$ of $F_\infty$ .
5. When is the sum of the first $n$ triangle areas for the first time bigger than $99.9\%$ of the area of $F_\infty$ ?

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Solution 6

It is:
1. The total areas of the objects (from left to right) are
  $\begin{array}{lll} F_1 &=& 3^2=9\\ F_2 &=& 9+ 3\cdot 1^2=9+3\\ F_3 &=& 9+3+9\cdot \left(\frac{1}{3}\right)^2 = 9+3+1\\ F_4 &=& 9+3+1 + 27\cdot \left(\frac{1}{9}\right)^2=9+3+1+\frac{1}{3}\\ ... &=& \\ \end{array}$
  So we see that the underlyung sequence of this series is
  $9,3,1,\frac{1}{3},...$
  Thus $q=\frac{1}{3}$ and therefore the total area is
  $F_\infty = 9\frac{1}{1-1/3}=\frac{27}{2}=13.5$
2. The total height in each generation is
  $\begin{array}{lll} H_1&=&3\\[0.5em] H_2&=&3+1\\[0.5em] H_3&=&3+1+\frac{1}{3}\\[0.5em] H_4&=&3+1+\frac{1}{3}+\frac{1}{9} ...&& \end{array}$
  Thus the underlying sequence of this series,
  $3,1,\frac{1}{3}, \frac{1}{9},...$
  is geometric with $q=\frac{1}{3}$ and it follows for the total height
  $H_\infty=3\frac{1}{1-\frac{1}{3}}=4.5$
  Now, find $n$ with
  $H_n =0.9999\cdot H_\infty$
  thus
  $3\frac{1-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}=0.9999\cdot 4.5$
  And it follows $n=8.38$ (apply logarithm), thus $n=9$ .
We have:
1. The side lengths $a_1=1, a_2=0.5, a_3=0.25, ...$ , form a geometric sequence with $q=0.5$ . It is $a_n=1\cdot 0.5^{n-1}$ Find $n$ with $a_n<0.0001$ , thus $0.5^{n-a}=0.0001$ $n = \frac{\ln(0.0001)}{\ln(0.5)}+1=14.28$ It follows $n=15$ .
2. The length $L_\infty$ of the bold spiral is $L_\infty = 1+0.5+0.25+...$ The sum forms a geometric sequence with $q=0.5$ , and because $-1<q<1$ if follows $L_\infty =1\cdot \frac{1}{1-0.5}=2$
3. Find $n$ with $L_n=1\cdot \frac{1-0.5^n}{1-0.5}>1.9999$ . We have $\begin{array}{lll} \frac{1-0.5^n}{1-0.5}&=1.9999\\ 0.5^n & = 0.00005\\ n & = \frac{\ln(0.00005)}{0.5}\\ & = 14.28 \end{array}$ Thus, it is $n=15$ .
4. The triangle areas have height (Pythagoras): $h_1=\frac{\sqrt{3}}{2}$ $h_2=\frac{\sqrt{3}}{4}$ $h_3=\frac{\sqrt{3}}{8}$ $...$ The triangle area are therefore $A_1=\frac{1}{2}\cdot\frac{\sqrt{3}}{2}\cdot 1=\frac{\sqrt{3}}{4}$ $A_2=\frac{1}{2}\cdot\frac{\sqrt{3}}{4}\cdot 0.5=\frac{\sqrt{3}}{16}$ $A_2=\frac{1}{2}\cdot\frac{\sqrt{3}}{8}\cdot 0.25=\frac{\sqrt{3}}{64}$ $...$ and we see that this is a geometric sequence with $q=0.25$ . Because of $-1<q<1$ if follows $A_\infty=\frac{\sqrt{3}}{4}\cdot\frac{1}{1-0.25}=0.577...$
5. $S_n=\frac{\sqrt{3}}{4}\cdot \frac{1-0.25^n}{1-0.25}$ . Find $n$ with $S_n>0.999\cdot F_\infty$ $\frac{\sqrt{3}}{4}\cdot \frac{1-0.25^n}{1-0.25} = 0.999\cdot \frac{\sqrt{3}}{4}\cdot\frac{1}{1-0.25}$ $0.25^n=0.001$ $n = \frac{\ln(0.001)}{\ln(0.25)}=4.983$ This it is $n=5$ .