3.2 The Summation (or Sigma) Notation

3.2.1 Definition

 

To facilitate the writing of lengthy sums, a shorthand notation, called summation notation or sigma notation is used.
The capital Greek letter sigma, \(\Sigma\), (equivalent to the Latin S) is used to denote summation as follows: let \(f\) be a function defined on \(\{1, 2, \dots , n\}\), then
\[\sum_{i=1}^{n}f(i)= f(1) + f(2) + \cdots + f(n).\]
For example,
\[\sum _{i=1}^5 i^2 = 1^2+2^2+3^3+4^2+5^5,\]
that is, the symbol \(\displaystyle \sum _{i=1}^5 i^2\) denotes the sum of the numbers \(i^2\) obtained by letting \(i\) run from 1 to 5. \(i\) is called the index of summation, and is a “dummy variable,” as it can be replaced by any other letter. For example,
\[\sum _{i=1}^4 \frac{i}{i+1}=\sum _{j=1}^4 \frac{j}{j+1}=\sum _{k=1}^4 \frac{k}{k+1}=\sum _{n=1}^4\frac{n}{n+1} =\]
\[= \frac{1}{1+1}+\frac{2}{2+1}+\frac{3}{3+1}+\frac{4}{4+1} = \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}.\]
Using this notation, we can write
\[R(f,5)=f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+f\left(x_3\right)\Delta x+f\left(x_4\right)\Delta x+f\left(x_5\right)\Delta x=\sum _{i=1}^5 f\left(x_i\right)\Delta x,\]
where \(x_1\), \(x_2\), \(\dots\), \(x_5\) represent the right-endpoints of the intervals.

On Your Own 1. Verify the numerical value of each sum
(a) \(\displaystyle \sum _{i=1}^3 (i^2+2i) = 26\);   (b) \(\displaystyle \sum _{k=1}^4\frac{1}{k}= \frac{25}{12}\).

Example 1. (Sums of Consecutive Odd Numbers.) Consider the sum \(\displaystyle \sum _{i=1}^k (2i-1)\).

(a) Compute the sum for \(k=1,2,3,4\); (b) conjecture a general formula for any \(k \in \mathbb{N}\).

Solution. (a) Denote by \(\displaystyle s_k=\sum _{i=1}^k (2i-1)\), then
\[ s_1=\sum _{i=1}^1 (2i-1)=2(1)-1=1, \;\;\; s_2=\sum _{i=1}^2 (2i-1)=( 2(1)-1) +( 2(2)-1)=1+3=4,\]
similarly,
\[s_3=\sum _{i=1}^3 (2i-1)=1+3+5=9,\;\; s_4=\sum _{i=1}^4 (2i-1)=1+3+5+7=16.\]

 

(b) Note the pattern:
\[s_1=1,\;\; s_2=4=2^2,\;\; s_3=9=3^2,\;\; s_4=16=4^2\,\dots .\]
Thus, we can conjecture that
\[s_k=\sum _{i=1}^k (2i-1)=k^2,\]
that is, the sum of the first \(k\) odd numbers is equal to \(k^2\).

On Your Own 2. (Sums of Consecutive Even Numbers.)

(a) Compute \(\displaystyle \sum _{i=1}^k 2i\) for \(k=1,2,3,4\);

(b) conjecture a general formula for any natural number \(k\).

 

Example 2. In Chapter 1, Proposition 2, we proved the following formula:
\[a^n – b^n = (a – b)\left(a^{n-1} + a^{n-2}b + a^{n-3} b^2+ \cdots + a^2b^{n-3}+ab^{n-2}+b^{n-1}\right).\]
Using sigma notation, this formula can be rewritten as follows:
\[a^n – b^n = (a – b)\sum_{i=1}^{n} a^{n-i}b^{i-1}.\]

3.2.2 Properties of Summations

 

Basic Properties

 

Proposition 1. Let \(a_i\), \(b_i\), \(i=1,\dots , n\) and \(c\) be real numbers, then
\begin{eqnarray}
\sum _{i=1}^n \left(a_i \pm b_i\right) &= &\sum _{i=1}^n a_i \pm \sum _{i=1}^n b_i, \label{ch03-02-eq01}\tag{3.1} \\
\sum _{i=1}^n c a_i &= &c\sum _{i=1}^n a_i.\label{ch03-02-eq02}\tag{3.2}
\end{eqnarray}

Proof.
\begin{align*}
\sum _{i=1}^n \left(a_i + b_i\right)& =(a_1+b_1)+(a_2+b_2)+(a_3+b_3)\cdots + (a_{n-1}+b_{n-1})+(a_n+b_n) = \\
&\underset{?}{=} (a_1+a_2+a_3+ \cdots + a_{n-1} + a_n ) + (b_1+a_2+b_3+ \cdots + b_{n-1} + b_n ) =\\
&=\sum _{i=1}^n a_i + \sum _{i=1}^n b_i.
\end{align*}
The other proofs are similar.

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Remark. The above properties extend to more terms. For example,
\[
\sum _{i=1}^n \left(r a_i + s b_i +t c_i – u d_i\right) =
r \sum _{i=1}^n a_i +
s \sum _{i=1}^n b_i +
t \sum _{i=1}^n c_i –
u \sum _{i=1}^n d_i.
\]

 

Telescoping Sums

 

We will be encountering the following special type of summation.

Definition 1.A sum \(\displaystyle \sum _{i=1}^n b_i\), with \(b_i= a_{i+1}-a_{i}\), \(a_i\in \mathbb{R}\), is called a telescoping sum}.

A formula to evaluate a telescoping sum is readily available.

Proposition 2. Assume that \(\displaystyle \sum _{i=1}^n b_i\) is telescoping with \(b_i= a_{i+1}-a_{i}\), \(a_i\in \mathbb{R}\). Then \(\displaystyle \sum _{i=1}^n b_i=a_{n+1}-a_1\)

Proof.
Since \(b_i= a_{i+1}-a_{i}\), then
\[
\sum _{i=1}^n b_i=
\left(a_2 – a_1\right) +
\left(a_3 – a_2\right) +
\left(a_4 – a_3\right) + \cdots +
\left(a_{n} – a_{n-1}\right) +
\left(a_{n+1} – a_n \right) = -a_1+a_{n+1}.
\]

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Example 1.[(Sums of Consecutive Odd Numbers) In Example 1 in 3.2.1, we conjectured that \(\displaystyle \sum _{i=1}^k (2i-1) = k^2\). Here we prove its validity by means of the formula for a telescoping sum.

(1) Note that \(2i-1 = i^2-(i-1)^2\).

This is immediate since \(i^2-(i-1)^2 = i^2-(i^2-2i+1)=2i-1\).

(2) Thus, we can conclude that the sum \(\displaystyle \sum _{i=1}^k (2i-1) = \sum _{i=1}^k(i^2-(i-1)^2)\) is a telescoping sum, with \(b_i=2i-1\), \(a_i=i^2\), and \(a_{i-1} =(i-1)^2\), thus
\[ \sum _{i=1}^k (2i-1) = \sum _{i=1}^k(i^2-(i-1)^2) = k^2-(1-1)^2 =k^2. \]

Note. We will use a similar procedure when proving other summation formulas below. Thus, make sure you understand this example very well.

 

Summation Formulas

 

The proposition below deals with sums of powers of consecutive natural numbers. There is a general procedure to obtain the formulas for such sums. Make sure you understand this general procedure.

Proposition 3.

(1) \(\displaystyle \sum _{i=1}^n 1=\underbrace{1+1+1+\text{…}+1+1}_n=n\).

(2) \(\displaystyle \sum _{i=1}^n i=1+2+3+\text{…}+(n-1)+n=\frac{n(n+1)}{2}\).

(3) \(\displaystyle \sum _{i=1}^n i^2=1^2+2^2+3^2+\text{…}+(n-1)^2+n^2=\frac{n(n+1)(2n+1)}{6}\).

(4) \(\displaystyle \sum _{i=1}^n i^3=1^3+2^3+3^3+\text{…}+(n-1)^3+n^3=\left(\frac{n(n+1)}{2}\right)^2\).

Proof. (1) This is immediate.

(2) Let \(\displaystyle S=\sum _{i=1}^n i\). First, write the sum \(S\) twice as shown below. Then add vertically down first, and then horizontally.
\[\begin{array}{cccccccccccc}
S & = & 1 & + & 2 & + & \text{…} & + & (n-1) & + & n & \\
S & = & n & + & (n-1) & + & \text{…} & + & 2 & + & 1 & \vphantom{\displaystyle \frac{A}{A^A}} \\ \hline
2S & = & (n+1) & + & (n+1) & + & \text{…} & + & (n+1) & + &(n+1) & \vphantom{\displaystyle \frac{A}{A^A}}
\end{array}\]
Then, \(\displaystyle 2S = n(n+1) \Longrightarrow S = \frac{n(n+1)}{2}\).

(3) The proof involves a telescoping sum, as in Example 1 in 3.2.1. Set \(\displaystyle S=\sum _{i=1}^n i^2\). To find a formula for \(S\), we are going to compute the sum \(\displaystyle \sum _{i=1}^n \left((i+1)^3-i^3\right)\) in two different ways.

(i) Note that \(\displaystyle \sum _{i=1}^n \left((i+1)^3-i^3\right)\) is a telescoping sum with \(b_i=(i+1)^3-i^3\). Thus, by Proposition 1 (2), we have
\[\sum _{i=1}^n \left((i+1)^3-i^3\right) = (n+1)^3-1^3= n^3+3n^2+3n .\]

 

(ii) Expanding first the expression \((i+1)^3-i^3\), and then adding, we have,
\begin{align*}
\displaystyle \sum _{i=1}^n \left((i+1)^3-i^3\right)
&\displaystyle =\sum _{i=1}^n (i^3+3i^2+3i+1-i^3)=\sum _{i=1}^n (3i^2+3i+1)=\\
&\displaystyle \underset{?}{=}3\sum _{i=1}^n i^2+3\sum _{i=1}^n i+\sum _{i=1}^n1
\underset{?}{=}
3S+3\frac{n(n+1)}{2}+n \underset{?}{=} 3S+\frac{3}{2}n^2+\frac{5}{2}n.
\end{align*}

 

It follows that
\[n^3+3n^2+3n+1-1=3S+\frac{3}{2}n^2+\frac{5}{2}n\Longleftrightarrow 3S=n^3+\frac{3}{2}n^2+\frac{n}{2}.\]
Finally, note that the last expression can be factored and simplify as follows
\[n^3+\frac{3}{2}n^2+\frac{n}{2}=\frac{n}{2}\left(2n^2+3n+1\right)=\frac{n(n+1)(2n+1)}{2}.\]
This proves part (3).

(4) The proof parallels the proof in (3). Set \(\displaystyle S=\sum _{i=1}^n i^3\).
Once again we are going to evaluate the sum \(\displaystyle \sum _{i=1}^n \left((i+1)^4-i^4\right)\) in two different ways.
First, the sum is telescoping, thus,
\[
\displaystyle \sum _{i=1}^n \left((i+1)^4-i^4\right)=(n+1)^4-1^4 =n^4+4n^3+6n^2+4n.
\]
Make sure to have clear why this is valid.

Second, we expand the expression in the sum, and then add.
\begin{align*}
\displaystyle \sum _{i=1}^n \left((i+1)^4-i^4\right)
&\displaystyle =\sum _{i=1}^n (i^4+4i^3+6i^2+4i+1-i^4)=\sum _{i=1}^n (4i^3+6i^2+4i+1)\\
&\displaystyle =4\sum _{i=1}^n i^3+6\sum _{i=1}^n i^2+4\sum _{i=1}^n i+\sum _{i=1}^n 1\\
&\displaystyle =4S+6\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}+n.
\end{align*}
It follows that
\[
(n+1)^4-1^4=4S+6\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}+n
\Longleftrightarrow \]
\[\Longleftrightarrow n^4+4n^3+6n^2+4n+1-1=4S+n(n+1)(2n+1)+2n(n+1)+n\Longleftrightarrow\]
\[\Longleftrightarrow n^4+4n^3+6n^2+4n=4S+n(2n^2+3n+1+2n+3)\Longleftrightarrow\]
\[\Longleftrightarrow n(n^3+4n^2+6n+4-2n^2-5n-4)=4S
\Longleftrightarrow n(n^3+2n^2+n)=4S\Longleftrightarrow\]
\[\Longleftrightarrow n(n(n+1)^2)=4S\Longleftrightarrow S=\frac{n^2(n+1)^2}{4}=\left(\frac{n(n+1)}{2}\right)^2.\]

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Note. Make sure that you are able to perform the steps of the above proofs on your own.

Example 2. Use the summation formulas in Proposition 3 to evaluate the sums.

(a) \(\displaystyle \sum_{i=1}^{10}(3i^2-4i+7)\); nbsp; (b) \(\displaystyle \sum_{i=1}^{n}(7i^3-5i^2-3i+2)\).

Solution.
(a)
\begin{align*}\sum_{i=1}^{10}(3i^2-4i+7) &
\underset{\;\;(3.1)\;\;}{=} \sum_{i=1}^{10}3i^2-\sum_{i=1}^{10}4i+\sum_{i=1}^{10}7
\underset{(3.2)}{=}3\sum_{i=1}^{10}i^2-4\sum_{i=1}^{10}i+7\sum_{i=1}^{10}1=\\
&\underset{\text{Prop.}3}{=}
3\frac{10(10+1)(2(10)+1)}{6} -4\frac{10(10+1)}{2} + 7(10) =\\
&\underset{\hphantom{\text{Prop.}\ref{ch03-02-prop03}}}{=} (55)(21) -220 + 70 = 1155-150 =1005.
\end{align*}

 

(b)
\begin{align*}
\sum_{i=1}^{n}(7i^3-5i^2-3i+2)
&\underset{\;\;(3.1)\;\;}{=}
\sum_{i=1}^{n}7i^3-\sum_{i=1}^{n}5i^2-\sum_{i=1}^{n}3i+\sum_{i=1}^{n}2=\\
&\underset{\;\;(3.2)\;\;}{=}
7\sum_{i=1}^{n}i^3-5\sum_{i=1}^{n}i^2-3\sum_{i=1}^{n}i + 2\sum_{i=1}^{n}1=\\
&\underset{\text{Prop.}3}{=}
7\left(\frac{n(n+1)}{2}\right)^2-5\frac{n(n+1)(2n+1)}{6}-3\frac{n(n+1)}{2} + 2n =\\
&\underset{\hphantom{\text{Prop.}3}}{=}
\frac{7}{4} n^2(n+1)^2-\frac{5}{6}n(n+1)(2n+1)-\frac{3}{2}n(n+1) + 2n.
\end{align*}

 

Remark. In part (b) we have not combined the different sums, nor have expanded the products. It really is not necessary to do so, and in future computations, we will see the value of having the answers in factored form. Thus, in the exercises, leave your answers in a similar form.