# Express Some Series in Sigma Notation

Sigma notation

Sigma notation is a method used to write out a long sum in a concise way. In this unit we look at ways of using sigma notation, and establish some useful rules.
In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature.
After reading this post and other posts in the child-category of sigma notation, you should be able to:
• expand a sum given in sigma notation into an explicit sum;
• write an explicit sum in sigma notation Where there is an obvious pattern to the individual terms;
• use rules to manipulate sums expressed in sigma notation.

Contents
● Introduction
● Some examples
● Writing a long sum in sigma notation
● Rules for use with sigma notation

Introduction
Sigma notation, Σ, provides a concise and convenient way of writing long sums. This leaflet explains how.
For example, the sum 1+2+3+4+5+⋯+10+11+12 can be written very concisely using the capital Greek letter Σ as The Σ stands for a sum, in this case the sum of all the values of k as k ranges through all Whole numbers from 1 to 12. Note that the lower—most and upper—most values of k are written at the bottom and top of the sigma sign respectively. You may also see this written as ∑k=12k=1 k, or even as ∑k=12k=1 k.

Sigma notation is a concise and convenient way to represent long sums. For example, we often wish to sum a number of terms such as

1+2+3+4+5

or
1+4+9+16+25+36

Where there is an obvious pattern to the numbers involved. The first of these is the sum of the first five whole numbers, and the second is the sum of the first six square numbers. More generally, if we take a sequence of numbers u1, u2, u3, ⋯, un then we can write the sum of these numbers as
u1+u2+u3+⋯+un.

A shorter way of writing this is to let ur represent the general term of the sequence and put Here, the symbol Σ is the Greek capital letter Sigma corresponding to our letter ‘S’, and refers to the initial letter of the word ‘Sum’. So this expression means the sum of all the terms ur where r takes the values from 1 to n. We can also write to mean the sum of all the terms ur where r takes the values from a to b. In such a sum, a is called the lower limit and b the upper limit.

Key Point: The sum u1+u2+u3+⋯+un is written in sigma notation as Sigma Notation is used to summarize a sum of several terms. 1. Start with r=0 (or whatever is below the sigma). Replace all of the r’s in the expression with r=0 02 2. Add 1 to r (it’s now r=1). Replace all of the r’s in the expression with r=1 02 12 3. Keep adding 1 to r and repeat until r is 5 (or whatever number is above the sigma). 02 12 22 32 42 52 4. Add all the terms together. 02+12+22+32+42+52

● Some Examples: ● Writing each long sum in sigma notation

Writing a long sum in sigma notation
Suppose that we are given a long sum and we want to express it in sigma notation. How should we do this?

Let us take the two sums we started with. If we want to write the sum

1+2+3+4+5

in sigma notation, we notice that the general term is just k and that there are 5 terms, so we would write To write the second sum
1+4+9+16+25+36

in sigma notation, we notice that the general term is k2 and that there are 6 terms, so We would write Example 1. We often see sums represented with summation notation. For example the sum of Gauss’ story may be represented as: where the symbol Σ stands for “summation”, i is known as an index, and the numbers 1 and 100 are the bounds of summation.
You try one! What sum is represented by ∑10i=2 j? (Answer: 2+3+4+⋯+9+10)

Ex2. Write the following series in sigma notation.
(a) 1+4+9+16+25
(b) 5t+6t+7t+8t+9t+10t
(d) 5+5+5+5+5+5
Solution: Ex3. Express each of the following in sigma notation:
(a) 1/1+½+⅓+¼+⅕
(b) -1+2-3+4-5+6-⋯+20
(c) (x1μ)2+(x2μ)2+(x3μ)2+(x4μ)2
Solution: Ex4. Express 1/1+½+⅓+¼ concisely using sigma notation.
Solution:
Each term takes the form 1/k Where k varies from 1 to 4. In sigma notation We could Write this as The sum

x1+x2+x3+x4+⋯+x19+x20

can be written _______________

There is nothing special about using the letter k. For example

n=7n=1 n2 stands for 12+22+32+42+52+62+72.

We can also use a little trick to alternate the signs of the numbers between + and -. Note that (-1)2=1, (-1)3=-1 and so on.

Ex5. Write the following sum in sigma notation. This one is a little more complicated. We’ll worry about the signs later, first we’ll deal with the numbers themselves. Do you notice a pattern in the terms? Sure, we get to the next term by dividing by 2. That is: If we call our index variable k, then k should go from 0 to 7, and the numbers themselves are just ½k. Now we need to deal with the signs.
We say above that (-1)k will alternate between +1 and -1. That is, if we multiply our terms from above by (-1)k, they will alternate between + and -. We are starting with k=0, so (-1)0=+1 will give us the alternation starting at the sign we want. Ex6. Write the sum

-1+½-⅓+¼-…+1/100

in sigma notation.
Solution
In this example, the first term -1 can also be Written as a fraction -1/1. We also notice that the signs of the terms alternate, with a minus sign for the odd—numbered terms and a plus sign for the even—numbered terms. So we can take care of the sign by using (-1)k, which is -1 when k is odd, and +1 when k is even. We can therefore write the sum as We can now see that k-th term is (-1)k 1/k, and that there are 100 terms, so We would Write the sum in sigma notation as Key Point: To write a sum in sigma notation, try to find a formula involving a variable is k where the first term can be obtained by setting k=1, the second term by k=2, and so on.

Ex7. Write the sum; ½-⅔+¾-⅘+⋯+99/100 in sigma notation. Solution
Solution:
We notice that each term’s numerator and denominator are consecutive integers, so they take on the absolute value of k/(k+1) or any equivalent form. We also notice that the signs of the terms alternate and that we have 99 terms. To take care of the sign, we use some power of (-1) that will start with a positive value. If we use (-1)k, the first term will be negative, so we can use (-1)(k+1) instead. We can, therefore, write the sum as Ex8: Write the expression in sigma notation.
• notice that we are adding fractions with a numerator of 1 and denominators starting with 1 in the first term and then increasing by 3 in each subsequent term;
• i.e. the denominator can be represented by 3k+1 for k=0, 1, ⋯, n;
• so we can Write this sum as Ex9. Write the expression in sigma notation.
• here the powers of x are even numbers which can be represented by 2k for k=0, 1, ⋯, n;
• the denominators are also even numbers but with factorials;
• so we can write this sum as Ex10. Express in summation notation:
Solution: Observe that 31=3, 32=9, 33=27, 34=81 and 35=343, also (-1)n=1 if n is even and (-1)n=-1 if n odd. Thus Ex11. Express in summation notation:
Solution: Observe that 21=2, 22=4, 23=8, 24=16 and 25=32, and that 31=3, 32=9, 33=27, 34=81, 35=343. Thus Ex12. Write the expression 3+6+9+12+⋯+60 using sigma notation.
• Notice that we are adding multiples of 3;
• then this sum can be written as ∑20n=1 3n. (Alternatively, you may index with i or k, etc.)
We can also use sigma notation when we have variables in our terms.
Ex13. Write the expression 3x+6x2+9x3+12x4+⋯+60x20 in sigma notation.
• The coefficients are successive multiples of 3, while the exponents on the x—term go up by 1 each time;
• then this sum can be Written as ∑20i=1 3ixi

Ex14. Write the following sums using summation notation. Solution:
The key to writing these sums with summation notation is to find the pattern of the terms.
(a) The terms of the sum 1, 3, 5, etc. , form an arithmetic sequence with first term a=1 and common difference d=2. We get a formula for the nth term of the series, the formula an=a+(n-1)d. Thus, the explicit nth term formula of the series is an=1+(n-1)2=2n-1, n≥1. At this stage, we have the formula for the terms, namely 2n-1, and the lower limit of the summation, n=1. To finish the problem, we need to determine the upper limit of the summation. In other words, we need to determine which value of n produces the term 117. Setting an=117, we get 2n-1=117 or n=59. Our final answer is (b) We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives ‘—’ with the numerators to get The numerators, 1, -1, etc. can be described by the geometric sequence((1)) cn=(-1)(n-1) for n≥1, while the denominators are given by the arithmetic sequence((2)) dn=n for n≥1. Hence, we get the formula an=(-1)(n-1)/n for our terms, and we find the lower and upper limits of summation to be n=1 and n=117, respectively. Thus _____
((1)) This is indeed a geometric sequence with first term a=1 and common ratio r=-1.
((2)) It is an arithmetic sequence with first term a=1 and common difference d=1.

(c) A formula for the nth term is a=9/10n for n≥1. This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the n-1 zeros to the right of the decimal point before the 9, we need a denominator of 10n. Hence, n is the upper limit of summation. Since n is used in the limits of the summation, we need to choose a different letter for the index of summation. (To see why, try writing the summation using ‘n’ as the index.) We choose k and get The following theorem presents some general properties of summation notation. While we shall not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover, there is much to be learned by thinking about why the properties hold. We invite the reader to prove these results. To get started, remember, “When in doubt, write it out!”

● Rules for use with sigma notation can be read at post Sigma Notation.

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