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Sums

Sums and products · Chapter 1 · Lecture notes (PDF) · Exercises: Sums · Slides (PDF) · App: Sums and products

1. Definition

Definition 1: sum

Given \(n\) values \(a_j \in \R\) with \(j \in \{1,2,\dots,n\}\), the sum

\[ a_1 + a_2 + {\rm \dots} + a_n \]

can be indicated in compact form with the sum symbol:

\[ \sum_{j=1}^n a_j \]

which reads: “sum for \(j\) from \(1\) to \(n\) of \(a_j\)”. The symbol \(j\) is called sum index.

  • The sum symbol is therefore very useful when the terms \(a_j\) are explicitly defined as a function of the sum index \(j\).

Example 1: sums

\[\begin{align*} \sum_{j=1}^{10} \frac{1}{j} &~~=~~ 1 +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{5} +\frac{1}{6} +\frac{1}{7} +\frac{1}{8} +\frac{1}{9} +\frac{1}{10} \\[2ex] \sum_{j=3}^{n} j^2 &~~=~~ 3^2 +4^2 +5^2 + {\rm \dots} +n^2 \end{align*}\]
  • The sum index is a dummy index. This means that if \(j\) is replaced with \(i\), \(k\) or any other index in all its occurrences the value of the sum does not change.

Example 2: dummy index

We have:

\[ \sum_{j=1}^n j^2 ~~=~~ \sum_{i=1}^n i^2 ~~=~~ \sum_{k=1}^{n} k^2 \]

as the two symbols indicate the sum of the squares of the first \(n\) natural numbers (without zero). On the contrary, we have:

\[ \sum_{j=1}^n j^2 ~~\neq~~ \sum_{j=1}^{m} j^2 \]

as the two symbols indicate the sum, respectively, of the first \(n\) and the first \(m\) squares of the natural numbers (without zero). Clearly, if \(n \neq m\), the result is different.

2. Properties

Observation 1: sum with constant term

For every natural number \(n\ge 1\) and real number \(c\), we have:

\[\begin{equation} \label{P2} \sum_{j=1}^n c = n \; c \end{equation}\]
Proof

We have:

\[ \underbrace{c\: + c + {\rm \dots} + c\:}_{=\sum_{j=1}^n c, {\rm ~~~} c {\rm ~summed~up~} n {\rm ~times}} = n \: c \]

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Observation 2: product by a constant

For every sum \(\sum_{j=1}^n a_j\) and real number \(c\), we have:

\[\begin{equation} \label{P1} \sum_{j=1}^n (c \; a_j) = c \: \sum_{j=1}^n a_j \end{equation}\]
Proof

From the distributive property, we have:

\[ \underbrace{c\: a_1 + c\: a_2 + {\rm \dots} + c\: a_n}_{=\sum_{j=1}^n (c \; a_j)} = \underbrace{c \: (a_1+a_2+{\rm \dots}+a_n)}_{= c \: \sum_{j=1}^n a_j} \]

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Observation 3: union of sums

For every pair of sums \(\sum_{j=1}^n a_j\) and \(\sum_{j=1}^n b_j\), we have:

\[\begin{equation} \label{P3} \sum_{j=1}^n a_j + \sum_{j=1}^n b_j = \sum_{j=1}^n (a_j + b_j) \end{equation}\]
Proof

We have:

\[ \underbrace{a_1 + a_2 + {\rm \dots} + a_n + b_1 + b_2 + {\rm \dots} + b_n}_{=\sum_{j=1}^n a_j + \sum_{j=1}^n b_j} = \underbrace{a_1 + b_1 +a_2 + b_2+{\rm \dots}+a_n + b_n}_{=\sum_{j=1}^n (a_j + b_j) } \]

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Observation 4: decomposition

For every sum \(\sum_{j=1}^{n+m} a_j\), we have:

\[\begin{align} \label{P4} \sum_{j=1}^{n+m} a_j &= \sum_{j=1}^n a_j + \sum_{j=n+1}^{n+m} a_j \end{align}\]

Observation 5: index translation

For every sum \(\sum_{j=1}^{n} a_j\) and natural number \(m \ge 1\), we have:

\[\begin{align} \label{P5} \sum_{j=1}^n a_j &= \sum_{j=1+m}^{n+m} a_{j-m} = \sum_{j=1-m}^{n-m} a_{j+m} \end{align}\]

Observation 6: index reflexion

For every sum \(\sum_{j=1}^{n} a_j\), we have:

\[\begin{align} \label{P6} \sum_{j=1}^n a_j &= \sum_{j=1}^n a_{n-j+1} = \sum_{j=0}^{n-1} a_{n-j} \end{align}\]
Proof

The three properties are simply different notations and/or arrangements of the terms in the sums. □

3. Some important sums

Observation 7: sum of the first \(n\) positive natural numbers

For every natural number \(n \ge 1\), we have:

\[\begin{equation} \sum_{j=1}^n j = \frac{n^2 +n}{2} \end{equation}\]
Proof

We have:

\[\begin{align*} \sum_{j=1}^n j &= \frac{1}{2} \left( \sum_{j=1}^n j + \sum_{j=1}^n j \right)\\[2ex] & = \frac{1}{2} \left( \sum_{j=1}^n j + \sum_{j=1}^n \big( n-j+1 \big) \right)\\[2ex] & = \frac{1}{2} \; \sum_{j=1}^n \big(j+ n-j+1 \big)\\[2ex] &= \frac{1}{2} \; \sum_{j=1}^n \big(n +1 \big) = \frac{n \: (n+1)}{2} = \frac{n^2 +n}{2} \end{align*}\]

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Observation 8: sum of the first \(n\) positive even numbers

For every natural number \(n \ge 1\), we have:

\[\begin{equation} \sum_{j=1}^n 2\:j = n^2 + n \end{equation}\]
Proof

We have:

\[\begin{align*} \sum_{j=1}^n 2\:j &= 2\:\sum_{j=1}^n j = 2 \left(\frac{n^2 + n}{2} \right) = n^2 + n \end{align*}\]

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Observation 9: sum of the first \(n\) odd numbers

For every natural number \(n \ge 1\), we have:

\[ \sum_{j=0}^{n-1} (2\:j+1) = n^2 \quad {\rm ~~or~~~equivalently~~} \quad \sum_{j=1}^n (2\:j-1) = n^2 \]
Proof

We have:

\[\begin{align*} \sum_{j=0}^{n-1} (2\: j+1) &= 2\:\sum_{j=0}^{n-1} j + {\sum_{j=0}^{n-1} 1} = 2\:\sum_{j=1}^{n} (j-1) + {\sum_{j=1}^{n} 1}\\[2ex] & = 2\:\left(\sum_{j=1}^n j - \sum_{j=1}^n 1 \right)+ n \\[2ex] & = 2\:\left( \frac{n^2 + n}{2} - n \right)+ n = n^2 + n - 2\:n + n = n^2 \\[5ex] \sum_{j=1}^n (2\:j-1) &= 2\:\sum_{j=1}^n j - {\sum_{j=1}^n 1}\\[2ex] & = 2\:\left( \frac{n^2 + n}{2} \right)- n = n^2 + n -n = n^2 \end{align*}\]

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Try it in the lab

the same computation step by step: change the matrix and watch the steps change.

Exercises and lab