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A34 |||| APPENDIX E SIGMA NOTATION

E SIGMA NOTATION

A convenient way of writing sums uses the Greek letter (capital sigma, corresponding to our letter S) and is called sigma notation.

This tells us to end with i=n.

	This tells us			n
	This tells us			μai
	to add.
	to add.
	This tells us to			i m


	start with i=m.
	start with i=m.

1 DEFINITION	If am, am 1, . . . , an are real numbers and m and n are integers
such that m n,	then
	n
	ai am am 1 am 2 an 1 an
	i m

With function notation, Deﬁnition 1 can be written as

f i f m f m 1 f m 2 f n 1 f n

i m

Thus the symbol ni m indicates a summation in which the letter i (called the index of summation) takes on consecutive integer values beginning with m and ending with n, that is, m, m 1, . . . , n. Other letters can also be used as the index of summation.

EXAMPLE 1

	4
(a)	i 2 12 22 32 42 30
	i 1
	n
(b)	i 3 4 5 n 1 n
	i 3
	5
(c)	2j 20 21 22 23 24 25 63
	j 0
	n	1			1			1				1
(d)		1		1	1			1				1
(d)		k		1								n
	k 1	k			2		3					n
	3	i 1			1 1				2 1				3 1				1	1	13
(e)																0
(e)		i	2	3	1	2	3			2	3		3	2	3	0	7	6	42
	i 1	i		3	1		3		2		3		3		3		7	6	42
	4
(f)	2 2 2 2 2 8																			M

i 1

EXAMPLE 2 Write the sum 23 33 n3 in sigma notation.

SOLUTION There is no unique way of writing a sum in sigma notation. We could write

	n
	23 33 n3 i 3
	i 2
	n 1
or	23 33 n3	j 1 3
	j 1
	n 2
or	23 33 n3	k 2 3	M

k 0

The following theorem gives three simple rules for working with sigma notation.

APPENDIX E SIGMA NOTATION |||| A35

2	THEOREM	If c is any constant (that is, it does not depend on i ), then
	n	n	n	n	n
(a)	cai c ai		(b)	ai bi ai	bi
	i m	i m	i m	i m	i m
	n	n	n
(c)	ai bi ai bi
	i m	i m	i m

PROOF To see why these rules are true, all we have to do is write both sides in expanded form. Rule (a) is just the distributive property of real numbers:

cam cam 1 can c am am 1 an

Rule (b) follows from the associative and commutative properties:

am bm am 1 bm 1 an bn


	am am 1 an bm bm 1 bn
Rule (c) is proved similarly.				M
	n
EXAMPLE 3 Find 1.
	i 1
	n
SOLUTION	1 1 1 1 n			M
	i 1
	n terms
EXAMPLE 4 Prove the formula for the sum of the ﬁrst n positive integers:
	n
	i 1 2 3 n	n n 1

	i 1		2

SOLUTION This formula can be proved by mathematical induction (see page 77) or by the following method used by the German mathematician Karl Friedrich Gauss (1777–1855) when he was ten years old.

Write the sum S twice, once in the usual order and once in reverse order:

S 1 2 3 n 1 n

S n n 1 n 2 2 1

Adding all columns vertically, we get

2S n 1 n 1 n 1 n 1 n 1

On the right side there are n terms, each of which is n 1, so

2S n n 1 or S n n 1 2

EXAMPLE 5 Prove the formula for the sum of the squares of the ﬁrst n positive integers:

i 2	12 22 32 n2 n n 1 2n 1
n
i 1	6

A36 |||| APPENDIX E SIGMA NOTATION

Most terms cancel in pairs.

N PRINCIPLE OF MATHEMATICAL INDUCTION

Let Sn be a statement involving the positive integer n. Suppose that

1. S1 is true.

SOLUTION 1 Let S be the desired sum. We start with the telescoping sum (or collapsing sum):

n
1 i 3 i3 23 13 33 23 43 33 n 1 3 n3
i 1
	n 1 3 13 n3 3n2 3n
On the other hand, using Theorem 2 and Examples 3 and 4, we have
n			n					n	n	n
	1 i 3 i 3 3i 2 3i 1 3 i 2 3								i	1
i 1			i 1					i 1	i 1 i 1
			3S 3		n n 1			n 3S 23 n2 25 n
			3S 3					n 3S 23 n2 25 n
						2
Thus we have
		n3 3n2 3n 3S 23 n2 25 n
Solving this equation for S, we obtain
	3S n3 23 n2 21 n
or		S	2n3 3n2 n				n n 1 2n 1
or		S	6				6
			6				6
SOLUTION 2 Let Sn be the given formula.
1. S1 is true because			12	1 1 1 2 1 1
1. S1 is true because			12
						6

2. If Sk is true, then Sk 1 is true.

Then Sn is true for all positive integers n.

N See pages 77 and 80 for a more thorough discussion of mathematical induction.

2. Assume that Sk is true; that is,
12 22 32 k2		k k 1 2k 1
12 22 32 k2
			6
Then
12 22 32 k 1 2 12 22 32 k2 k 1 2
	k k 1 2k 1			k 1 2
				k 1 2
			6
k 1			k 2k 1 6 k 1
k 1
			6
k 1			2k2 7k 6
k 1			6
			6
	k 1 k 2 2k 3

			6
	k 1 k 1 1 2 k 1 1

			6
So Sk 1 is true.
By the Principle of Mathematical Induction, Sn is true for all n.					M

N The type of calculation in Example 7 arises in Chapter 5 when we compute areas.

APPENDIX E SIGMA NOTATION |||| A37

We list the results of Examples 3, 4, and 5 together with a similar result for cubes (see Exercises 37–40) as Theorem 3. These formulas are needed for ﬁnding areas and evaluating integrals in Chapter 5.

THEOREM Let c be a constant and n a positive integer. Then

(a)

1 n

(b)

c nc

(c)

i 1

(d)

i 1

n n 1 2n 1

i n n 1

i 2

i 1

n n

(e)

i 3

EXAMPLE 6 Evaluate i 4i 2 3 .

i 1

SOLUTION

Using Theorems 2 and 3, we have

i 4i 2 3 4i 3 3i 4 i 3 3

i 1

n n

n n 1

n n 1 2n n 1 3

n n 1 2n2 2n 3

n n

n l i 1

EXAMPLE 7 Find lim

1 .

SOLUTION

n l i 1 n n

n l i 1 n

lim

i 2

n l n

i 1

lim

n n 1 2n 1

n l

2 n

lim

2n 1

n l

lim

21 1 1 2 3 4

A38 |||| APPENDIX E SIGMA NOTATION

E EXERCISES

1–10 Write the sum in expanded form.

	5			6	1
1.	si		2.		1
1.	si		2.		i 1
	i 1			i 1	i 1
	6			6
3.	3i		4.	i 3
	i 4			i 4
	4	2k 1		8
5.		2k 1	6.	x k
5.		2k 1	6.	x k
	k 0	2k 1		k 5
	n			n 3
7.	i 10		8.	j 2
	i 1			j n
	n 1			n
9.		1 j	10.	f xi xi
	j 0			i 1

11–20 Write the sum in sigma notation.

11.1 2 3 4 10

12.s3 s4 s5 s6 s7

13.12 23 34 45 1920

14.37 48 59 106 2327

15.2 4 6 8 2n

16.1 3 5 7 2n 1

17.1 2 4 8 16 32

18.11 14 19 161 251 361

19.x x 2 x 3 x n

20.1 x x 2 x 3 1 nx n

21–35 Find the value of the sum.

	8		6
21.	3i 2	22.	i i 2
	i 4		i 3
	6		8
23.	3 j 1	24.	cos k
	j 1		k 0
	20		100
25.	1 n	26.	4
	n 1		i 1
	4		4
27.	2i i 2	28.	23 i
	i 0		i 2
	n		n
29.	2i	30.	2 5i
	i 1		i 1
	n		n
31.	i 2 3i 4	32.	3 2i 2
	i 1		i 1
	n		n
33.	i 1 i 2	34.	i i 1 i 2
	i 1		i 1

35. i 3 i 2

i 1

36. Find the number n such that i 78.

i 1

37.Prove formula (b) of Theorem 3.

38.Prove formula (e) of Theorem 3 using mathematical induction.

39.Prove formula (e) of Theorem 3 using a method similar to that of Example 5, Solution 1 [start with 1 i 4 i 4 .

40.Prove formula (e) of Theorem 3 using the following method published by Abu Bekr Mohammed ibn Alhusain Alkarchi in about AD 1010. The ﬁgure shows a square ABCD in which

sides AB and AD have been divided into segments of lengths 1, 2, 3, . . . , n. Thus the side of the square has length n n 1 2 so the area is n n 1 2 2. But the area is also the sum of the

areas of the n “gnomons” G1, G2, . . . , Gn shown in the ﬁgure. Show that the area of Gi is i 3 and conclude that formula (e) is true.

nGn

							.									.	.
							.							.		.
							.							.

						5						G∞
						4
						4				G¢
						3			G£
						2				G™
						2				G™
						1
						A1 2 3				4		5			...				n		B
						A1 2 3				4		5							n		B
41. Evaluate each telescoping sum.
	n											100
	(a) i 4 i 1 4											(b)		5i 5i 1
	i 1												i 1
	99	1				1								n
	(c)	1				1						(d) ai ai 1
	(c)				i 1							(d) ai ai 1
	i 3 i				i 1								i 1
42. Prove the generalized triangle inequality:
									n			n
									i 1			i 1
									ai						ai
43– 46 Find the limit.													n l i 1 n							n
	n l i 1 n		n										n l i 1 n							n
	n	1		i		2												n	1		i	3
43.	lim		n									44.		lim									1
	n l i 1 n		n							n
	n	2				2i	3				2i
45.	lim							5

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