 # 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 ai as i goes from 1 to n].

## Example 1 See the terms of the given series.

43, 53, 63, ... 193

So ai = i3.

The first term is 43.
And the last term is 193.

So i goes from 4 to 19.

ai = i3
i goes from 4 to 19.

So the given series is

the sum of i3
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

a1 = 1
d = 2

So an = 1 + (n - 1)⋅2.

Arithmetic sequences

(n - 1)⋅2 = 2n - 2

1 - 2 = -1

So an = 2n - 1.

There are 7 terms.

So n goes from 1 to 7.

an = 2n - 1
n goes from 1 to 7.

So the given series is

the sum of [2n - 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 ak = 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.

ak = 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 ai = 3i - 1

As i increases +1,
ai increases +3.

So the given summation shows
an arithmetic series
whose d = 3.

Arithmetic sequences

ai = 3i - 1

So a1 = 3⋅1 - 1.

So a1 = 2.

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

So n = 10.

The given summation is the arithmetic series:
S10.

a1 = 2
d = 3
n = 10

So S10 = (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 ak = 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 ak = 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.

ak = 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.

ax is the power of i: ix.

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 i1,
group four powers:

[i1 + i2 + i3 + i4] + [i5 + i6 + i7 + i8] + ....

The quotient was 24.
So there are 24 of these [groups].

The remainder was 3.
So write the last three terms:

+ [i97 + i98 + i99].

i = i
i2 = -1
i3 = -i
i4 = +1

i5 = i4⋅1 + 1 = i
i6 = i4⋅1 + 2 = -1
i7 = i4⋅1 + 3 = -i
i8 = i4⋅2 + 0 = +1

As you can see,
[i, -1, -i, +1] are repeating.

So
i97 = i,
i98 = -1,
i99 = -i.

This is why you divide 99 by .

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.