Infinite Geometric Series

An infinite series means
a series when n goes to infinity.

S = a₁ + a₂ + a₃ + ... + a_n + ...

For an infinite geometric series,
if the common ratio r is -1 < r < 1
(|r| < 1),
then the sum S goes to this value.

S = a/[1 - r]

S: Infinite geometric series
a: First term
r: Common ratio

If |r| < 1 (-1 < r < 1),
as n increases,
a_n goes to 0.
So the sum S_n = [a(1 - rⁿ)]/[1 - r]
goes to
S = [a(1 - 0)]/[1 - r]
= a/[1 - r].

Write the terms of the given series
as a_n = ar^{n - 1} form.

Geometric Sequence

Write the first term 1.

This is a.

The second term is +1/2.

So write
plus,
a, 1
times,
(+1/2)/1, 1/2
to the first:
+1⋅(1/2)¹.

The third term is +1/4.

a = 1

So write +1⋅(1/2)².

The fourth term is +1/8.

a = 1

So write +1⋅(1/2)³.

Write + ... .

So
1 + 1/2 + 1/4 + 1/8 + ...
= 1 + 1⋅(1/2)¹ + 1⋅(1/2)² + 1⋅(1/2)³ + ... .

a = 1
r = 1/2
n goes to ∞.

So the sum is
1/[1 - 1/2].

Multiply 2
to both of the numerator and the denominator.

1⋅2 = 2

[1 - 1/2]⋅2 = 2 - 1

Common Monomial Factor

2 - 1 = 1

2/1 = 2

So 2 is the answer.

Write the terms of the given series
as a_n = ar^{n - 1} form.

Write the first term 10.

This is a.

The second term is -20/3.

So write
plus,
a, 10
times,
(-20/3)/10, -2/3
to the first:
+10⋅(-2/3)¹.

The third term is +40/9.

a = 10

So write +10⋅(-2/3)².

The fourth term is -80/27.

a = 10

So write +10⋅(-2/3)³.

Write + ... .

So
10 - 20/3 + 40/9 - 80/27 + ...
= 10 + 10⋅(-2/3)¹ + 10⋅(-2/3)² + 10⋅(-2/3)³ + ... .

a = 10
r = -2/3
n goes to ∞.

So the sum is
10/[1 - (-2/3)].

-(-2/3) = +2/3

Multiply 3
to both of the numerator and the denominator.

10⋅3 = 30

(1 + 2/3)⋅3 = 3 + 2

3 + 2 = 5

30/5 = 6

So 6 is the answer.

See the sigma notation.
n goes from 0.

So the first term, a₁, is
when n = 0.

So a₁ = 8⋅(1/7)⁰.

(1/7)⁰ = 1

Zero Exponent

8⋅1 = 8

So a₁ = 8.

See the term in the sigma:
8⋅(1/7)ⁿ.

As n increases,
(1/7) is multiplied:
8⋅(1/7)⁰, 8⋅(1/7)¹, 8⋅(1/7)², ... .

So the given summation
is an infinite geometric series.

So r = 1/7.

a₁ = a = 8
r = 1/7
n goes to ∞.

So the sum is
8/[1 - 1/7].

Multiply 7
to both of the numerator and the denominator.

8⋅7 = 56

(1 - 7)⋅7 = 7 - 1

7 - 1 = 6

Reduce the numerator 56 to, 56/2, 28
and reduce 6 to, 6/2, 3.

So 28/3 is the answer.