# Sigma Notation

How to write and solve a series in sigma notation: definition, examples, and their solutions.

## Definition

Sigma notation is a simple way

to write the sum of a sequence (= series).

[Σ] is the greek capital letter [sigma].

The given summation means (and is read as)

[the sum of *a*_{i} as *i* goes from 1 to *n*].

## Example 1

See the terms of the given series.

4^{3}, 5^{3}, 6^{3}, ... 19^{3}

So *a*_{i} = *i*^{3}.

The first term is 4^{3}.

And the last term is 19^{3}.

So *i* goes from 4 to 19.

*a*_{i} = *i*^{3}*i* goes from 4 to 19.

So the given series is

the sum of *i*^{3}

as *i* goes from 4 to 19.

## Example 2

See the terms of the given series.

3 - 1 = 2

5 - 3 = 2

7 - 5 = 2

It's an arithmetic series.

Arithmetic series*a*_{1} = 1*d* = 2

So *a*_{n} = 1 + (*n* - 1)⋅2.

Arithmetic sequences

(*n* - 1)⋅2 = 2*n* - 2

1 - 2 = -1

So *a*_{n} = 2*n* - 1.

There are 7 terms.

So *n* goes from 1 to 7.

*a*_{n} = 2*n* - 1*n* goes from 1 to 7.

So the given series is

the sum of [2*n* - 1]

as *n* goes from 1 to 7.

## Example 3

See the terms of the given series.

1/[1⋅(1 + 1)]

1/[2⋅(2 + 1)]

1/[3⋅(3 + 1)]

...

1/[99⋅(99 + 1)]

So *a*_{k} = 1/[*k*(*k* + 1)].

The first term is 1/[1⋅(1 + 1)].

And the last term is 1/[99⋅(99 + 1)].

So *k* goes from 1 to 99.

*a*_{k} = 1/[*k*(*k* + 1)]*k* goes from 1 to 99.

So the given series is

the sum of 1/[*k*(*k* + 1)]

as *k* goes from 1 to 99.

## Example 4

*a*_{i} = 3*i* - 1

As *i* increases +1,*a*_{i} increases +3.

So the given summation shows

an arithmetic series

whose *d* = 3.

Arithmetic sequences

*a*_{i} = 3*i* - 1

So *a*_{1} = 3⋅1 - 1.

So *a*_{1} = 2.

See the given summation.*i* goes from 1 to 10.

So *n* = 10.

The given summation is the arithmetic series:*S*_{10}.*a*_{1} = 2*d* = 3*n* = 10

So *S*_{10} = (10/2)[2⋅2 + (10 - 1)⋅3].

Arithmetic series

10/2 = 5

2⋅2 = 4

10 - 1 = 9

9⋅3 = 27

4 + 7 = 31

5⋅31 = 155

So (given) = 155.

## Example 5

See the terms of the given series.

1/[1⋅(1 + 1)]

1/[2⋅(2 + 1)]

1/[3⋅(3 + 1)]

...

1/[99⋅(99 + 1)]

So *a*_{k} = 1/[*k*(*k* + 1)].

(*k* + 1) - *k* = 1

The result is a constant.

So use the partial fraction decompostion formula.

So 1/[*k*(*k* + 1)] = 1/[(*k* + 1) - *k*] ⋅ (1/*k* - 1/(*k* + 1)).

1/*AB* = 1/(*B* - *A*)(1/*A* - 1/*B*)

1/[(*k* + 1) - *k*] = 1/1

So *a*_{k} = 1/*k* - 1/(*k* + 1).

The first term is 1/[1⋅(1 + 1)].

And the last term is 1/[99⋅(99 + 1)].

So *k* goes from 1 to 99.

*a*_{k} = 1/*k* - 1/(*k* + 1)*k* goes from 1 to 99.

So the given series is

the sum of [1/*k* - 1/(*k* + 1)]

as *k* goes from 1 to 99.

Expand the summation.

Start from *k* = 1 to *k* = 99.

So (given)

= (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/99 - 1/100).

Cancel -1/2 and 1/2.

Cancel -1/3 and 1/3.

Then -1/4 will be cancelled with the next 1/4.

And 1/99 will be cancelled with the last -1/99.

Then the remaining terms are

1/1 and -1/100.

When cancelling like this,

the remaining terms remain symmetrically.

1/1 = 100/100

100/100 - 1/100 = 99/100

So (given) = 99/100.

## Example 6

See the given summation.*a*_{x} is the power of *i*: *i*^{x}.

Imaginary number *i*

In this case,

start from finding the number of the terms.*x* goes from 1 to 99.

So there are 99 terms.

Divide 99 by 4.

The quotient is 24.

And the remainder is 3.

Expand the given summation.

Starting from *i*^{1},

group four powers:

[*i*^{1} + *i*^{2} + *i*^{3} + *i*^{4}] + [*i*^{5} + *i*^{6} + *i*^{7} + *i*^{8}] + ....

The quotient was 24.

So there are 24 of these [groups].

The remainder was 3.

So write the last three terms:

+ [*i*^{97} + *i*^{98} + *i*^{99}].

*i* = *i**i*^{2} = -1*i*^{3} = -*i**i*^{4} = +1*i*^{5} = *i*^{4⋅1 + 1} = *i**i*^{6} = *i*^{4⋅1 + 2} = -1*i*^{7} = *i*^{4⋅1 + 3} = -*i**i*^{8} = *i*^{4⋅2 + 0} = +1

As you can see,

[*i*, -1, -*i*, +1] are repeating.

So*i*^{97} = *i*,*i*^{98} = -1,*i*^{99} = -*i*.

This is why you divide 99 by [4].

Powers of *i*

*i* - 1 - *i* + 1 = 0

So each group is 0.

For the last three terms,*i* - 1 - *i* = -1.

So (given) = -1.