Sequences and Series

Definition of Sequence

A sequence is a function with domain the natural numbers (infinite sequence) or some subset of the natural numbers from 1 up to some larger number (finite sequence). The notation $ f(n) = a_{n} $ is used to denote range elements of the function: the $ a_{1}, a_{2}, a_{3}, $ are called the first, second, third, etc. terms of the sequence, and $ a_{n} $ is referred to as the nth term. The independent variable n is referred to as the index. Unless otherwise specified, a sequence is assumed to be an infinite sequence.

EXAMPLE 41.1 Write the first four terms of the sequence specified by $ a_{n}=2n $ .

\[ a_{1}=2\cdot1,a_{2}=2\cdot2,a_{3}=2\cdot3,a_{4}=2\cdot4.The sequence would be written 2\cdot1,2\cdot2,2\cdot3,2\cdot4,\ldots or 2,4,6,8,\ldots. \]

EXAMPLE 41.2 Write the first four terms of the sequence specified by $ a_{n}=(-1)^{n} $

$ a_{1}=(-1){1},a_{2}=(-1){2},a_{3}=(-1){3},a_{4}=(-1){4} $ . The sequence would be written $ (-1){1},(-1){2},(-1){3},(-1){4},$ or $ -1,1,-1,1,$ .

Finding the nth Term of a Sequence

Given the first few terms of a sequence, a common exercise is to determine the nth term, that is, a formula which generates all the terms. In fact, such a formula is not uniquely determined, but in many cases a simple one can be developed.

EXAMPLE 41.3 Find a formula for the nth term of the sequence 1, 4, 9, 16, …

Notice that the terms are all perfect squares, and the sequence could be written $ 1^{2} $ , $ 2^{2} $ , $ 3^{2} $ , $ 4^{2} $ , $ $ .

Thus the nth term of the sequence can be given as $ a_{n}=n^{2} $ .

Recursive Defined Sequence

A sequence is defined recursively by specifying the first term and defining later terms with respect to earlier terms.

EXAMPLE 41.4 Write the first four terms of the sequence defined by $ a_{1}=3 $ , $ a_{n}=a_{n-1}+7 $ , n>1.

For n = 1, $ a_{1} = 3 $

\[ For n=2,a_{2}=a_{2-1}+7=a_{1}+7=3+7=10 \]

\[ For n=3,a_{3}=a_{3-1}+7=a_{2}+7=10+7=17 \]

\[ For n=4,a_{_{4}}=a_{_{4-1}}+7=a_{_{3}}+7=17+7=24 \]

The sequence can be written 3, 10, 17, 24, $ $ .

Definition of Series

A series is the indicated sum of the terms of a sequence. Thus if $ a_{1}, a_{2}, a_{3}, , a_{m} $ are the m terms of a finite sequence, then associated with the sequence is the series given by $ a_{1} + a_{2} + a_{3} + + a_{m} $ . Series are often written using the summation notation:

\[ a_{1}+a_{2}+a_{3}+\cdots+a_{m}=\sum_{k=1}^{m}a_{k} \]

Here $ $ is called the summation symbol, and k is called the index of summation or just the index. The right-hand side of this definition is read, “the sum of the $ a_{k} $ , with k going from 1 to m.”

EXAMPLE 41.5 Write in expanded form: $ _{k=1}{5}\frac{1}{k{2}} $

Replace k, in turn, with the integers from 1 to 5 and add the results:

\[ \sum_{k=1}^{5}\frac{1}{k^{2}}=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25} \]

I nfinite Series

The sum of all the terms of an infinite sequence is referred to as an infinite series, and is indicated by the symbol:

\[ \sum_{k=1}^{\infty}a_{k} \]

Infinite series in general are discussed in calculus courses; a special case (infinite geometric series) is treated in Chapter 43.

Factorial Symbol

A useful definition is the factorial symbol. For natural numbers n, n! (pronounced n factorial) is defined as the product of the natural numbers from 1 up to n. Then

\[ 1!=1\qquad2!=1\cdot2=2\qquad3!=1\cdot2\cdot3=6\qquad4!=1\cdot2\cdot3\cdot4=24 \]

and so on. Separately, 0! is defined to equal 1.

SOLVED PROBLEMS

41.1. Write the first four terms of the sequences specified by

\[ \begin{array}{l}(a)a_{_{n}}=2n-1;(b)b_{_{n}}=6-4n;(c)c_{_{n}}=2^{n};(d)d_{_{n}}=3(-2)^{n}\end{array} \]

  1. $ a_{1}=2-1=1 $ , $ a_{2}=2-1=3 $ , $ a_{3}=2-1=5 $ , $ a_{4}=2-1=7 $ . The sequence would be written 1, 3, 5, 7, $ $ .

  2. $ b_{1}=6-4=2 $ , $ b_{2}=6-4=-2 $ , $ b_{3}=6-4=-6 $ , $ b_{4}=6-4=-10 $ . The sequence would be written 2, -2, -6, -10, $ $ .

  3. $ c_{1}=2^{1}=2 $ , $ c_{2}=2^{2}=4 $ , $ c_{3}=2^{3}=8 $ , $ c_{4}=2^{4}=16 $ . The sequence would be written 2, 4, 8, 16, . . .

  4. $ d_{1}=3(-2)^{1}=-6, d_{2}=3(-2)^{2}=12, d_{3}=3(-2)^{3}=-24, d_{4}=3(-2)^{4}=48 $ . The sequence would be written -6, 12, -24, 48, $ $ .

41.2. Write the first four terms of the sequences specified by

\[ \begin{aligned}(a)a_{n}=\frac{1}{3n+1};(b)b_{n}=\frac{n^{2}}{3n-2};(c)c_{n}=\sin\frac{\pi n}{4};(d)d_{n}=\frac{(-1)^{n}\sqrt{n}}{(n+1)(n+2)}\end{aligned} \]

\[ \left(a\right)a_{1}=\frac{1}{3\cdot1+1}=\frac{1}{4},a_{2}=\frac{1}{3\cdot2+1}=\frac{1}{7},a_{3}=\frac{1}{3\cdot3+1}=\frac{1}{10},a_{4}=\frac{1}{3\cdot4+1}=\frac{1}{13} \]

\[ \left(\mathrm{b}\right)b_{1}=\frac{1^{2}}{3\cdot1-2}=1,b_{2}=\frac{2^{2}}{3\cdot2-2}=1,b_{3}=\frac{3^{2}}{3\cdot3-2}=\frac{9}{7},b_{4}=\frac{4^{2}}{3\cdot4-2}=\frac{8}{5} \]

\[ (c)c_{1}=\sin\frac{\pi\cdot1}{4}=\frac{1}{\sqrt{2}},c_{2}=\sin\frac{\pi\cdot2}{4}=1,c_{3}=\sin\frac{\pi\cdot3}{4}=\frac{1}{\sqrt{2}},c_{4}=\sin\frac{\pi\cdot4}{4}=0 \]

\[ \left(\mathrm{d}\right)d_{1}=\frac{(-1)^{1}\sqrt{1}}{(1+1)(1+2)}=-\frac{1}{6},d_{2}=\frac{(-1)^{2}\sqrt{2}}{(2+1)(2+2)}=\frac{\sqrt{2}}{12},d_{3}=\frac{(-1)^{3}\sqrt{3}}{(3+1)(3+2)}=-\frac{\sqrt{3}}{20}, \]

\[ d_{4}=\frac{(-1)^{4}\sqrt{4}}{(4+1)(4+2)}=\frac{\sqrt{4}}{30}=\frac{1}{15} \]

41.3. Write the tenth term of each of the sequences in the previous problem.

\[ a_{10}=\frac{1}{3\cdot10+1}=\frac{1}{31} \]

\[ b_{10}=\frac{10^{2}}{3\cdot10-2}=\frac{25}{7} \]

\[ c_{10}=\sin\frac{\pi\cdot10}{4}=1 \]

\[ d_{10}=\frac{(-1)^{10}\sqrt{10}}{(10+1)(10+2)}=\frac{\sqrt{10}}{132} \]

41.4. Write the first four terms of the following recursively defined sequences:

\[ \left(a\right)a_{1}=1,a_{n}=na_{n-1},n>1;\left(b\right)a_{1}=1,a_{n}=a_{n-1}+2,n>1;\left(c\right)a_{1}=12,a_{n}=\frac{a_{n-1}}{4},n>1 \]

\[ (a)For n=1,a_{1}=1 \]

  1. For n = 1, $ a_{1} = 1 $

\[ For n=2,a_{2}=2a_{2-1}=2a_{1}=2\cdot1=2 \]

\[ For n=2,a_{2}=a_{2-1}+2=a_{1}+2=1+2=3 \]

\[ For n=3,a_{3}=3a_{3-1}=3a_{2}=3\cdot2=6 \]

\[ For n=3,a_{3}=a_{3-1}+2=a_{2}+2=3+2=5 \]

\[ For n=4,a_{4}=4a_{4-1}=4a_{3}=4\cdot6=24 \]

\[ For n=4,a_{4}=a_{4-1}+2=a_{3}+2=5+2=7 \]

\[ (c)For n=1,a_{1}=12 \]

\[ For n=2,a_{2}=\frac{a_{2-1}}{4}=\frac{a_{1}}{4}=\frac{12}{4}=3 \]

\[ For n=3,a_{3}=\frac{a_{3-1}}{4}=\frac{a_{2}}{4}=\frac{3}{4} \]

\[ For n=4,a_{4}=\frac{a_{4-1}}{4}=\frac{a_{3}}{4}=\frac{3/4}{4}=\frac{3}{16} \]

41.5. The sequence defined by \(a_{1}=1\), \(a_{2}=1\), \(a_{n}=a_{n-1}+a_{n-2}\), \(n>2\), is called a Fibonacci sequence. Write the first six terms of this sequence.

\[ For n=1,a_{1}=1 \]

\[ For n=2,a_{2}=1 \]

\[ For n=3,a_{3}=a_{2}+a_{1}=1+1=2\qquad For n=4,a_{4}=a_{3}+a_{2}=2+1=3 \]

\[ For n=5,a_{5}=a_{4}+a_{3}=3+2=5\qquad For n=6,a_{6}=a_{5}+a_{4}=5+3=8 \]

The sequence would be written 1, 1, 2, 3, 5, 8, $ $ .

41.6. Find a formula for the nth term of a sequence whose first four terms are given by:

  1. Comparing the terms of the sequence with $ n = 1, 2, 3, 4, $ shows that the individual terms are each 2 times the index n of the term. Thus a possible formula would be $ a_{n} = 2n $ .

  2. The sequence can be written as $ , , , , $ ; thus, comparing the denominators with the previous sequence shows that each denominator is 1 less than 2, 4, 6, 8, $ $ , hence, can be written as 2n - 1. Thus a possible formula would be $ a_{n} = $ .

  3. The absolute values of the terms of the sequence are powers of 2, that is, $ 2^{0} $ , $ 2^{1} $ , $ 2^{2} $ , $ 2^{3} $ ; comparing this to $ n = 1 $ , $ 2 $ , $ 3 $ , $ 4 $ , $ $ suggests that the nth term has absolute value $ 2^{n-1} $ . The fact that the signs of the terms alternate can be represented (in more than one way) by successive powers of -1, for example, $ (-1)^{1} $ , $ (-1)^{2} $ , $ (-1)^{3} $ , $ (-1)^{4} $ , $ $ . Thus a possible formula would be $ a_{n} = (-1){n}2{n-1} $ .

  4. A pattern for the denominators can be found by comparing to the sequence 1, 4, 9, 16, $ $ of Example 41.3; since each denominator is 1 more than the corresponding term of this sequence, they can be represented by $ n^{2} + 1 $ . The numerators are equal to the index of the terms; thus, a possible formula would be $ a_{n} = $ .

41.7. Write each series in expanded form:

\[ \begin{aligned}(a)\sum_{k=1}^{4}(6k+1);(b)\sum_{j=1}^{5}\frac{j}{j^{2}+1};(c)\sum_{j=3}^{20}(-1)^{j-1}(5j);(d)\sum_{k=1}^{p}\frac{k^{k}}{k!}\end{aligned} \]

  1. Replace k in the expression $ 6k + 1 $ , in turn, with each natural number from 1 to 4, and place an addition symbol between the results:

\[ \sum_{k=1}^{4}\left(6k+1\right)=\left(6\cdot1+1\right)+\left(6\cdot2+1\right)+\left(6\cdot3+1\right)+\left(6\cdot4+1\right)=7+13+19+25=64 \]

  1. Note that the letter j is used for the index here; in general, any variable letter may be used, but the letters i, j, k are the most common. Replace j in the expression after the summation symbol, in turn, with each natural number from 1 to 5, and place an addition symbol between the results.

\[ \sum_{j=1}^{5}\frac{j}{j^{2}+1}=\frac{1}{1^{2}+1}+\frac{2}{2^{2}+1}+\frac{3}{3^{2}+1}+\frac{4}{4^{2}+1}+\frac{5}{5^{2}+1}=\frac{1}{2}+\frac{2}{5}+\frac{3}{10}+\frac{4}{17}+\frac{5}{26} \]

In this context, it is not always necessary to complete the arithmetic; if desired, the addition can be performed to yield $ $ .

  1. Note that the index starts from 3; there is no requirement that a series must start from index 1. Replace j in the expression after the summation symbol, in turn, with each natural number from 3 to 20, and place an addition symbol between the results.

\[ \begin{aligned}\sum_{j=3}^{20}(-1)^{j-1}(5j)&=(-1)^{3-1}(5\cdot3)+(-1)^{4-1}(5\cdot4)+(-1)^{5-1}(5\cdot5)+\cdots+(-1)^{20-1}(5\cdot20)\\ &=15-20+25-\cdots-100\end{aligned} \]

If there are an unwieldy number of terms, as in this case, not all terms are written out explicitly; the three dots (ellipsis . . .) symbol is used.

  1. Here the variable on top of the summation symbol indicates that the number of terms is not explicitly stated. Write out the first few terms and the last term; use the ellipsis symbol.

\[ \sum_{k=1}^{p}\frac{k^{k}}{k!}=\frac{1^{1}}{1!}+\frac{2^{2}}{2!}+\frac{3^{3}}{3!}+\cdots+\frac{p^{p}}{p!} \]

41.8. Write the following series in expanded form:

\[ \begin{aligned}(a)\sum_{k=1}^{3}\frac{x^{k+1}}{k};(b)\sum_{k=1}^{5}(-1)^{k-1}x^{k};(c)\sum_{k=0}^{4}\frac{(-1)^{k}x^{k}}{k!}\end{aligned} \]

  1. Replace k in the expression after the summation symbol, in turn, with each natural number from 1 to 3, and place an addition symbol between the results.

\[ \sum_{k=1}^{3}\frac{x^{k+1}}{k}=\frac{x^{1+1}}{1}+\frac{x^{2+1}}{2}+\frac{x^{3+1}}{3}=x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{3} \]

  1. Replace k in the expression after the summation symbol, in turn, with each natural number from 1 to 5, and place an addition symbol between the results.

\[ \begin{aligned}\sum_{k=1}^{5}(-1)^{k-1}x^{k}&=(-1)^{1-1}x^{1}+(-1)^{2-1}x^{2}+(-1)^{3-1}x^{3}+(-1)^{4-1}x^{4}+(-1)^{5-1}x^{5}\\&=x-x^{2}+x^{3}-x^{4}+x^{5}\end{aligned} \]

  1. Replace k in the expression after the summation symbol, in turn, with each integer from 0 to 4, and place an addition symbol between the results.

\[ \begin{aligned}\sum_{k=0}^{4}\frac{(-1)^{k}x^{k}}{k!}&=\frac{(-1)^{0}x^{0}}{0!}+\frac{(-1)^{1}x^{1}}{1!}+\frac{(-1)^{2}x^{2}}{2!}+\frac{(-1)^{3}x^{3}}{3!}+\frac{(-1)^{4}x^{4}}{4!}\\&=\frac{1}{0!}-\frac{x}{1!}+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\frac{x^{4}}{4!}or1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+\frac{x^{4}}{24}\end{aligned} \]

41.9. Write the following series in summation notation:

\[ \begin{aligned}(a)3+6+9+12+15;(b)\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16};(c)4+\frac{4}{3}+\frac{4}{9}+\cdots+\frac{4}{729};(d)\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}\end{aligned} \]

  1. Comparing the terms of the sequence with $ k = 1, 2, 3, 4, $ shows that the individual terms are each 3 times the index k of the term. Thus a possible formula for the terms would be $ a_{k} = 3k $ ; there are 5 terms, hence the series can be written as $ _{k=1}^{5} 3k $ .

  2. Comparing the terms of the sequence with $ k = 1, 2, 3, 4, $ shows that the denominators are powers of 2: $ 2^{1}, 2^{2}, 2^{3}, $ . The fact that the signs of the terms alternate can be represented by successive powers of -1, for example, $ (-1)^{0}, (-1)^{1}, (-1)^{2}, $ . Thus a possible formula for the terms would be $ a_{k} = (-1){k-1}/2{k} $ ; there are 4 terms, hence the series can be written as $ _{k=1}^{4} (-1){k-1}/2{k} $ .

  3. Comparing the terms of the sequence with \(k=1,2,3,4,\ldots\) shows that the denominators are powers of \(3:3^{0},3^{1},3^{2},\ldots\). Thus a possible formula for the terms would be \(a_{k}=4/3^{k-1}\). Since the last term has denominator \(729=3^{6}\), setting \(6=k-1\) yields \(k=7\); there are 7 terms, hence the series can be written as \(\sum_{k=1}^{7}4/3^{k-1}\).

  4. Comparing the terms of the sequence with \(k=1,2,3,4,\ldots\) shows that the denominators are representable as factorials; a possible formula for the terms would be \(a_{k}=x^{k}/k!\); there are 4 terms, hence the series can be written as \(\sum_{k=1}^{4}x^{k}/k!\).

SUPPLEMENTARY PROBLEMS

41.10. Write the first four terms of the following sequences: (a) $ a_{n} = $ ; (b) $ a_{n} = $ ; (c) $ a_{n} = 5 - 2n $ ;

Ans. (a) $ , , , $ ; (b) $ , , , $ ; (c) 3, 1, -1, -3;

  1. 2, 0, 6, 0; (e) 5, 9, 17, 33; (f) 4, $ - $ , $ $ , $ - $

41.11. The sequence recursively defined by $ a_{n}= $ , with $ a_{1} $ chosen arbitrarily, may be used to approximate $ $ to any desired degree of accuracy. Find the first four terms of the sequence $ a_{1}=2 $ , $ a_{n}= $ , n>1 and compare to the calculator approximation for $ $ .

Ans. 2, 2.25, 2.23611, 2.236068, calculator: $ $

41.12. Find a formula for the nth term of a sequence whose first four terms are given by:

\[ Ans.\quad(a)a_{n}=3n+1;(b)a_{n}=(-1)^{n-1}(2n-1);(c)a_{n}=(-1)^{n-1}\frac{n+5}{2n+5};(d)a_{n}=\frac{x^{2n}}{(2n)!} \]

41.13. Write in expanded form: (a) $ {k=1}{4}\frac{(-2){k}}{k+1} $ ; (b) $ {k=3}{6}\frac{x{k}}{(k+1)!} $ ; (c) $ _{k=0}{3}\frac{x{k}}{(k+1)(k+3)} $

Ans. (a) $ +++ $ ; (b) $ +++ $ ; (c) $ +++ $

41.14. Write the following in summation notation:

\[ \left(a\right)\frac{1}{3}+\frac{2}{5}+\frac{3}{7}+\frac{4}{9};\left(b\right)x-2x^{2}+3x^{3}-4x^{4}+5x^{5}-6x^{6};\left(c\right)x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!} \]

\[ Ans.\quad(a)\sum_{k=1}^{4}\frac{k}{2k+1};(b)\sum_{k=1}^{6}(-1)^{k-1}kx^{k};(c)\sum_{k=1}^{4}\frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!} \]