Special Sequences and Series

Definition of Arithmetic Sequence

A sequence of numbers $ a_{n} $ is called an arithmetic sequence if successive terms differ by the same constant, called the common difference. Thus $ a_{n}-a_{n-1}=d $ and $ a_{n}=a_{n-1}+d $ for all terms of the sequence. It can be proved by mathematical induction that for any arithmetic sequence, $ a_{n}=a_{1}+(n-1)d $ .

Definition of Arithmetic Series

An arithmetic series is the indicated sum of the terms of a finite arithmetic sequence. The notation $ S_{n} $ is often used, thus, $ S_{n} = {k=1}^{n} a{k} $ . For an arithmetic series,

\[ S_{n}=\frac{n}{2}(a_{1}+a_{n})\quad S_{n}=\frac{n}{2}[2a_{1}+(n-1)d] \]

EXAMPLE 43.1 Write the first 6 terms of the arithmetic sequence 4, 9, $ $

Since the sequence is arithmetic, with $ a_{1}=4 $ and $ a_{2}=9 $ , the common difference d is given by $ a_{2}-a_{1}=9-4=5 $ . Thus, each term can be found from the previous term by adding 5, hence the first 6 terms are 4, 9, 14, 19, 24, 29.

EXAMPLE 43.2 Find the sum of the first 20 terms of the sequence of the previous example.

To find $ S_{20} $ , either of the formulas for an arithmetic series may be used. Since $ a_{1}=4 $ , n=20, and d=5 are known, the second formula is more convenient:

\[ S_{n}=\frac{n}{2}\left[2a_{1}+(n-1)d\right] \]

\[ S_{20}=\frac{20}{2}\left[2\cdot4+(20-1)5\right]=1030 \]

Definition of Geometric Sequence

A sequence of numbers $ a_{n} $ is called a geometric sequence if the quotient of successive terms is a constant, called the common ratio. Thus $ a_{n} a_{n-1} = r $ or $ a_{n} = ra_{n-1} $ for all terms of the sequence. It can be proved by mathematical induction that for any geometric sequence, $ a_{n} = a_{1}r^{n-1} $ .

Definition of Geometric Series

A geometric series is the indicated sum of the terms of a geometric sequence. For a geometric series with $ r $ ,

\[ S_{_{n}}=a_{_{1}}\frac{1-r^{n}}{1-r} \]

EXAMPLE 43.3 Write the first 6 terms of the geometric sequence 4, 6, . . . .

Since the sequence is geometric, with $ a_{1}=4 $ and $ a_{2}=6 $ , the common ratio r is given by $ a_{2}a_{1}=6=3/2 $ . Thus, each term can be found from the previous term by multiplying by 3/2, hence the first 6 terms are 4, 6, 9, 27/2, 81/4, 243/8.

EXAMPLE 43.4 Find the sum of the first 8 terms of the sequence of the previous example.

Use the sum formula with $ a_{1}=4 $ , n=8, and r=3/2:

\[ S_{_{n}}=a_{_{1}}\frac{1-r^{n}}{1-r} \]

\[ S_{8}=4\frac{1-(3/2)^{8}}{1-(3/2)}=\frac{6305}{32} \]

I nfinite Geometric Series

It is not possible to add up all the terms of an infinite geometric sequence. In fact, if $ |r| $ , the sum is not defined. However, it can be shown in calculus that if $ |r| < 1 $ , then the sum of all the terms, denoted by $ S_{} $ , is given by:

\[ S_{\infty}=\frac{a_{1}}{1-r} \]

EXAMPLE 43.5 Find the sum of all the terms of the geometric sequence 6, 4, . . . .

Since the sequence is geometric, with $ a_{1}=6 $ and $ a_{2}=4 $ , the common ratio r is given by $ a_{2}a_{1}=4=2/3 $ . Therefore $ S_{}===18 $ .

Series Identities

The following identities can be proved by mathematical induction:

\[ \begin{aligned}\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}=\sum_{k=1}^{n}(a_{k}+b_{k})\quad&\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{n}b_{k}=\sum_{k=1}^{n}(a_{k}-b_{k})\quad\sum_{k=1}^{n}ca_{k}=c\sum_{k=1}^{n}a_{k}\\\sum_{k=1}^{n}c=cn\quad&\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\quad\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}\\\sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4}\quad&\sum_{k=1}^{n}k^{4}=\frac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30}\end{aligned} \]

SOLVED PROBLEMS

43.1. Identify the following sequences as arithmetic, geometric, or neither.

1. 2, 4, 8, $ $ ; (b) $ $ , $ $ , $ $ , $ $ ; (c) 7, 5, 3, $ $ ; (d) $ $ , $ $ , $ $ , $ $

2. Since $ a_{2}-a_{1}=4-2=2 $ and $ a_{3}-a_{2}=8-4=4 $ , the sequence is not arithmetic. Since $ a_{2}/a_{1}=4/2=2 $ and $ a_{3}/a_{2}=8/4=2 $ , the sequence is geometric with a common ratio of 2.

3. Since $ a_{2}-a_{1}=-=- $ and $ a_{3}-a_{2}=-=- $ , the sequence is not arithmetic. Since $ a_{2}a_{1}== $ and $ a_{3}a_{2}== $ , the sequence is not geometric. Thus it is neither arithmetic nor geometric.

4. Since $ a_{2}-a_{1}=5-7=-2 $ and $ a_{3}-a_{2}=3-5=-2 $ , the sequence is arithmetic with a common difference of -2.

5. Since $ a_{2}-a_{1}=-=- $ and $ a_{3}-a_{2}=-=- $ , the sequence is not arithmetic. Since $ a_{2}a_{1}== $ and $ a_{3}a_{2}== $ , the sequence is geometric with a common ratio of $ $ .

43.2. Identify the following sequences as arithmetic, geometric, or neither.

\[ 3,\frac{15}{4},\frac{9}{2},\cdots;(b)\ln1,\ln2,\ln3,\cdots;(c)x^{-1},x^{-2},x^{-3},\cdots;(d)0.1,0.11,0.111,\cdots. \]

1. Since $ a_{2}-a_{1}=-3= $ and $ a_{3}-a_{2}=-= $ , the sequence is arithmetic with a common difference of $ $ .

2. Since $ a_{2}-a_{1}=-= $ and $ a_{3}-a_{2}=-= $ , the sequence is not arithmetic. Since $ a_{2}a_{1}=()() $ is not defined, the sequence is not

geometric. Thus it is neither arithmetic nor geometric.

3. Since $ a_{2}-a_{1}=x^{-2}-x{-1}= $ and $ a_{3}-a_{2}=x^{-3}-x{-2}= $ , the sequence is not arithmetic except in the special case x=1. Since $ a_{2}a_{1}=x^{-2}x^{-1}=x{-1} $ and $ a_{3}a_{2}=x^{-3}x^{-2}=x{-1} $ , the sequence is geometric with a common ratio of $ x^{-1} $ (except in the special case x=0).

4. Since \(a_{2}-a_{1}=0.11-0.1=0.01\) and \(a_{3}-a_{2}=0.111-0.11=0.001\), the sequence is not arithmetic. Since \(a_{2}\div a_{1}=0.11\div0.1=1.1\) and \(a_{3}\div a_{2}=0.111\div0.11\approx1.01\), the sequence is not geometric. Thus it is neither arithmetic nor geometric.

43.3. Prove that for an arithmetic sequence the nth term is given by $ a_{n}=a_{1}+(n-1)d $

An arithmetic sequence is defined by the relation $ a_{n}=a_{n-1}+d $ . Let $ P_{n} $ be the statement that $ a_{n}=a_{1}+(n-1)d $ and proceed by mathematical induction.

$ P_{1} $ is the statement $ a_{1}=a_{1}+(1-1)d $ .

$ P_{k} $ is the statement $ a_{k}=a_{1}+(k-1)d $ .

$ P_{k+1} $ is the statement $ a_{k+1} = a_{1} + [(k+1)-1]d $ , which can be rewritten as

\[ a_{k+1}=a_{1}+kd \]

Now $ P_{1} $ is obviously true. Assume the truth of $ P_{k} $ and note that by the definition of an arithmetic sequence, $ a_{k+1} = a_{k} + d $ . Therefore

\[ a_{k+1}=a_{k}+d=a_{1}+(k-1)d+d=a_{1}+kd. \]

But $ a_{k+1} = a_{1} + k d $ is precisely the statement $ P_{k+1} $ . Thus the truth of $ P_{k+1} $ follows from the truth of $ P_{k} $ . Thus, by the principle of mathematical induction, $ P_{n} $ holds for all n.

43.4. Given that the following sequences are arithmetic, find the common difference and write the next three terms and the nth term.

1. 2, 5, $ $ ; (b) 9, $ $ , $ $ ; (c) $ $ , $ $ , $ $

2. The common difference is 5 - 2 = 3. Each term is found by adding 3 to the previous term, hence the next three terms are 8, 11, 14. The nth term is found from $ a_{n} = a_{1} + (n - 1)d $ with $ a_{1} = 2 $ and d = 3; thus $ a_{n} = 2 + (n - 1)3 = 3n - 1 $ .

3. The common difference is $ - 9 = - $ . Each term is found by adding $ - $ to the previous term, hence the next three terms are 8, $ $ , 7. The nth term is found from $ a_{n} = a_{1} + (n - 1)d $ with $ a_{1} = 9 $ and $ d = - $ ; thus $ a_{n} = 9 + (n - 1)(-) = $ .

4. The common difference is $ - = $ . Each term is found by adding $ $ to the previous term, hence the next three terms are given by $ + = $ , $ + = $ , and $ + = $ . The nth term is found from $ a_{n} = a_{1} + (n - 1)d $ with $ a_{1} = $ and $ d = $ ; thus

$ a_{n}=+(n-1)=(n-1)=^{n-1}. $

43.5. Given that the following sequences are geometric, find the common ratio and write the next three terms and the nth term.

1. 5, 10, $ $ ; (b) 4, -2, $ $ ; (c) 0.03, 0.003, $ $

2. The common ratio is $ 10 = 2 $ . Each term is found by multiplying the previous term by 2, hence the next three terms are 20, 40, 80. The nth term is found from $ a_{n} = a_{1} r^{n-1} $ with $ a_{1} = 5 $ and r = 2; thus $ a_{n} = 5 ^{n-1} $ .

3. The common ratio is $ -2 = - $ . Each term is found by multiplying the previous term by $ - $ ; hence the next three terms are 1, $ - $ , $ $ . The nth term is found from $ a_{n} = a_{1} r^{n-1} $ with $ a_{1} = 4 $ and $ r = - $ ; thus $ a_{n} = 4 ( - )^{n-1} = $ .

4. The common ratio is $ 0.003 = 0.1 $ . Each term is found by multiplying the previous term by 0.1; hence the next three terms are 0.0003, 0.00003, 0.000003. The nth term is found from

\[ a_{_{n}}=a_{_{1}}r^{n-1}with a_{_{1}}=0.03and r=0.1,thus a_{_{n}}=0.03(0.1)^{n-1}=3\times10^{-2}\times10^{1-n}=\frac{3}{10^{n+1}}. \]

43.6. Derive the formulas $ S_{n} = (a_{1} + a_{n}) $ and $ S_{n} = [2a_{1} + (n - 1)d] $ for the value of an arithmetic series. To derive the first formula, write out the terms of $ S_{n} $ ;

\[ S_{n}=a_{1}+(a_{1}+d)+(a_{1}+2d)+\cdots+[a_{1}+(n-1)d] \]

Now write the terms in reverse order, noting that to begin with $ a_{n} $ , each term is found by subtracting d, the common difference, from the previous term.

\[ S_{_{n}}=a_{_{n}}+(a_{_{n}}-d)+(a_{_{n}}-2d)+\cdots+[a_{_{1}}-(n-1)d] \]

Adding these two identities, term by term, and noting that all terms involving \(d\) add to zero, yields:

\[ S_{n}+S_{n}=(a_{1}+a_{n})+(a_{1}+a_{n})+(a_{1}+a_{n})+\cdots+(a_{1}+a_{n}) \]

Since there are n identical terms on the right,

\[ \begin{aligned}2S_{_{n}}&=n(a_{_{1}}+a_{_{n}})\\S_{_{n}}&=\frac{n}{2}(a_{_{1}}+a_{_{n}})\end{aligned} \]

For the second formula, substitute $ a_{n}=a_{1}+(n-1)d $ into the above to obtain

\[ S_{n}=\frac{n}{2}[a_{1}+a_{1}+(n-1)d] \]

\[ S_{n}=\frac{n}{2}[2a_{1}+(n-1)d] \]

43.7. Find the sum of the first 10 terms of the arithmetic sequences given in Problem 43.4.

1. Here $ a_{1}=2 $ and $ a_{n}=3n-1 $ . For n=10,

\[ S_{10}=\frac{10}{2}[2+(3\cdot10-1)]=155 \]

2. Here $ a_{1}=9 $ and $ a_{n}= $ . For n=10,

\[ S_{10}=\frac{10}{2}\bigg[9+\frac{19-10}{2}\bigg]=\frac{135}{2} \]

3. Here $ a_{1} = $ and $ a_{n} = ^{n-1} $ . For n = 10,

\[ S_{10}=\frac{10}{2}[\ln1+\ln2^{10-1}]=5\ln2^{9}=45\ln2 \]

43.8. Derive the formula $ S_{n} = a_{1} $ for the value of a finite geometric series ( $ r $ ).

Write out the terms of $ S_{n} $ .

\[ S_{n}=a_{1}+a_{1}r+a_{1}r^{2}+\cdot\cdot\cdot+a_{1}r^{n-1} \]

Multiply both sides by r to obtain

\[ rS_{n}=a_{1}r+a_{1}r^{2}+a_{1}r^{3}+\cdot\cdot\cdot+a_{1}r^{n} \]

Subtracting these two identities, term by term, yields:

\[ \begin{aligned}S_{n}-rS_{n}&=a_{1}-a_{1}r^{n}\\S_{n}(1-r)&=a_{1}(1-r^{n})\end{aligned} \]

Assuming $ r $ , both sides may be divided by 1 - r to yield $ S_{n} = a $ . Note that if r = 1, then

\[ S_{n}=a_{1}+a_{1}+a_{1}+\cdots+a_{1}=na_{1} \]

43.9. Find the sum of the first 7 terms of the geometric sequences given in Problem 43.5.

1. Here $ a_{1}=5 $ and r=2. For n=7,

\[ S_{7}=5\bigg(\frac{1-2^{7}}{1-2}\bigg)=635 \]

2. Here $ a_{1}=4 $ and $ r=- $ . For n=7,

\[ S_{7}=4\left[\frac{1-\left(-\frac{1}{2}\right)^{7}}{1-\left(-\frac{1}{2}\right)}\right]=4\bigg(\frac{2^{7}+1}{2^{7}+2^{6}}\bigg)=\frac{129}{48} \]

3. Here clearly $ S_{7} = 0.03333333 $ . Using the formula is more cumbersome, but yields the same result.

43.10. Give a plausibility argument to justify the formula $ S_{} = $ for the sum of all the terms of an infinite geometric sequence, $ |r| < 1 $ .

First note that there are three possibilities: \(r=0,0<r<1\), and \(-1<r<0\). For the first case, the formula is clearly valid, since all terms after the first are zero; hence \(S_{\infty}=a_{1}+0=\frac{a_{1}}{1-0}\). For \(0<r<1\), consider the formula \(S_{n}=a_{1}\frac{1-r^{n}}{1-r}\) and let \(n\) increase beyond all bounds. Since for real \(n\), \(r^{n}\) is then an exponential decay function, as \(n\to\infty\), \(r^{n}\to0\). It seems plausible that this remains valid if \(n\) is restricted to integer values. Thus, as \(n\to\infty\), \(S_{n}\to a_{1}\frac{1}{1-r}\) and \(S_{\infty}=\frac{a_{1}}{1-r}\). A similar, but more cumbersome, argument can be given if \(-1<r<0\). A convincing proof is left for a calculus course.

43.11. Find the sum of all the terms of each geometric sequence given in Problem 43.5, or state that the sum is undefined.

1. Since r = 2, the sum of all the terms is not defined. The sequence is said to diverge.

2. Since $ r = - $ and $ a_{1} = 4 $ , $ S_{} = = $ .

3. Since r = 0.1 and $ a_{1} = 0.03 $ , $ S_{} = = $ .

43.12. Use mathematical induction to show that $ {j=1}^{{n}a_{j}+_{j=1}}{n}b{j}={j=1}^{n}(a{j}+b_{j}) $ holds for all positive integers n.

Let $ P_{n} $ be the above statement. Then

$ P_{1} $ is the statement $ {j=1}^{{1}a_{j}+_{j=1}}{1}b{j}={j=1}^{1}(a{j}+b_{j}). $

$ P_{k} $ is the statement $ {j=1}^{{k}a_{j}+_{j=1}}{k}b{j}={j=1}^{k}(a{j}+b_{j}). $

$ P_{k+1} $ is the statement $ {j=1}^{{k+1}a_{j}+_{j=1}}{k+1}b{j}={j=1}^{k+1}(a{j}+b_{j}). $

Now $ P_{1} $ is true, since it reduces to $ a_{1} + b_{1} = (a_{1} + b_{1}) $ . Assume the truth of $ P_{k} $ ; then

\[ a_{1}+a_{2}+\cdots+a_{k}+b_{1}+b_{2}+\cdots+b_{k}=(a_{1}+b_{1})+(a_{2}+b_{2})+\cdots+(a_{k}+b_{k}) \]

Add $ a_{k+1} + b_{k+1} $ to both sides, then

\[ \begin{aligned}a_{1}+a_{2}+\cdots+a_{k}+b_{1}+b_{2}+\cdots+b_{k}+a_{k+1}+b_{k+1}&=(a_{1}+b_{1})+(a_{2}+b_{2})+\cdots+(a_{k}+b_{k})\\&\quad+(a_{k+1}+b_{k+1})\end{aligned} \]

Rearranging the terms on the left side yields

\[ \begin{aligned}a_{1}+a_{2}+\cdots+a_{k}+a_{k+1}+b_{1}+b_{2}+\cdots+b_{k}+b_{k+1}&=(a_{1}+b_{1})+(a_{2}+b_{2})+\cdots+(a_{k}+b_{k})\\&\quad+(a_{k+1}+b_{k+1})\end{aligned} \]

Writing this in the summation notation, it becomes

\[ \sum_{j=1}^{k+1}a_{j}+\sum_{j=1}^{k+1}b_{j}=\sum_{j=1}^{k+1}(a_{j}+b_{j}) \]

But this is precisely the statement $ P_{k+1} $ . Thus the truth of $ P_{k+1} $ follows from the truth of $ P_{k} $ . Thus, by the principle of mathematical induction, $ P_{n} $ holds for all n.

43.13. Determine the seating capacity of a lecture hall if there are 32 rows of seats, with 18 seats in the first row, 21 seats in the second row, 24 seats in the third row, and so on.

The number of seats in each row forms an arithmetic sequence, with $ a_{1}=18 $ , d=21-18=3, and n=32. Use the second formula for the value of an arithmetic series:

\[ S_{n}=\frac{n}{2}[2a_{1}+(n-1)d] \]

\[ S_{32}=\frac{32}{2}[2\cdot18+(32-1)3]=2064 \]

43.14. A company buys a machine that is valued at $87,500 and depreciates it at the rate of 30% per year. What is the value of the machine at the end of 5 years?

Note that depreciation of 30% of the value of the machine means that at the end of each year the value is 70% of what it was at the beginning. Thus the value at the end of each year is a constant multiple of the value at the end of the previous year. Hence the values form a geometric sequence, with $ a_{1} = (0.7)(87500) $ (the value at the end of the first year), r = 0.7 and n = 5. Thus

\[ \begin{aligned}&a_{_{n}}=a_{_{1}}r^{n-1}\\&a_{_{5}}=(0.7)(87500)(0.7)^{5-1}\approx14706\\ \end{aligned} \]

The value of the machine is $14,706.

43.15. A ball is dropped from a height of 80 feet and bounces to three-fourths of its initial height. Assuming that this process continues indefinitely, find the total distance travelled by the ball before coming to rest.

Initially the ball travels 80 feet before hitting the ground. It then bounces up to a height of $ (80) $ and then back down this same distance. As this process repeats, the distance traveled can be written:

\[ 80+2\Big(\frac{3}{4}\Big)80+\frac{3}{4}\Big[2\Big(\frac{3}{4}\Big)80\Big]+\cdots \]

Except for the first term, this may be regarded as an infinite geometric series with $ a_{1}=2()80=120 $ and $ r= $ . Hence, if the process continues indefinitely, the entire distance travelled is given by

\[ 80+S_{_{\infty}}=80+\frac{120}{1-\frac{3}{4}}=560feet \]

SUPPLEMENTARY PROBLEMS

43.16. Are the following sequences arithmetic, geometric, or neither?

\[ \begin{array}{l}(a)\frac{3}{8},\frac{3}{2},6,\cdots;(b)\frac{3}{8},\frac{3}{4},\frac{9}{8},\cdots;(c)\frac{3}{4},\frac{3}{5},\frac{3}{6},\cdots;(d)\frac{3}{4},-\frac{3}{4},\frac{3}{4},-\frac{3}{4},\cdots;(e)\frac{3}{4},\frac{4}{5},\frac{5}{6},\cdots\end{array} \]

Ans. (a) geometric; (b) arithmetic; (c) neither; (d) geometric; (e) neither

43.17. For the following arithmetic sequences, state the common difference, and write the next three terms and the nth term: (a) $ , , $ ; (b) $ -8, -5, $ ; (c) $ , 3, $

Ans. (a) $ d = $ ; $ 1, , $ ; $ a_{n} = $ ; (b) d = 3; -2, 1, 4; $ a_{n} = 3n - 11 $ ;

\[ (c)d=2\pi;5\pi,7\pi,9\pi;a_{n}=(2n-1)\pi \]

43.18. Prove by mathematical induction: for a geometric sequence, the nth term is given by $ a_{n}=a_{1}r^{n-1} $

43.19. For the following geometric sequences, state the common ratio, and write the next three terms and the nth term: (a) $ $ , $ $ , $ $ ; (b) -5, 5, -5, $ $ ; (c) 1, 1.05, $ $

\[ Ans.\quad(a)r=8;6,48,384;a_{n}=3\cdot2^{3n-8};(b)r=-1;5,-5,5;a_{n}=5(-1)^{n}; \]

\[ \begin{array}{r l}{(\mathrm{c})r=1.05;(1.05)^{2},(1.05)^{3},(1.05)^{4};a_{n}=(1.05)^{n-1}}\end{array} \]

43.20. For the following geometric sequences, state the common ratio and find the sum of all terms, or state that the sum is undefined. (a) $ 4, , , $ ; (b) $ , -, , $ ; (c) 36, -12, 4, $ $ ; (d) 1, 0.95, $ $

Ans. (a) $ r = , S_{} = $ ; (b) r = -1, sum undefined; (c) $ r = -, S_{} = 27 $ ; (d) $ r = 0.95, S_{} = 20 $

43.21. Use mathematical induction to show that $ {k=1}^{{n}a_{k}-_{k=1}}{n}b{k}={k=1}^{n}(a{k}-b_{k}) $ and $ {k=1}^{{n}ca_{k}=c_{k=1}}{n}a{k} $ hold for all integers n.

43.22. Suppose that $0.01 were deposited into a bank account on the first day of June, $0.02 on the second day, $0.04 on the third day, and so on in a geometric sequence. (a) How much money would be deposited at this rate on June 30th? (b) How much money would be in the account after this last deposit?

Ans. (a) $5,368,709.12; (b) $10,737,418.23