Calculus II: Arithmetic series

1. Arithmetic series

An arithmetic series is a series in which each term is equal to the sum of its previous term and a fixed common difference.

That is a series of the form:

Σ an =	a1 + a2 + ... + an + ...
n → ∞

where:
a₂ = r + a₁
a₃ = r + a₂ = 2r + a₁
a₄ = 3r + a₁
... = ...
a_n = (n - 1)r + a₁

The sum of the first terms is:
s_n = a₁ + a₂ + ... + a_n
= a₁ + r + a₁ + 2r + a₁ + ... + (n - 1)r + a₁
= n a₁ + r (1 + 2 + 3 + ... + n-1)

Let's write:
z = 1 + 2 + 3 + ... + n-2 + n-1
so
z = n-1 + (n -2 ) + ( n - 3) + ... + 2 + 1
adding the two equations gives:
2 z = n ( n - 1)
so
z = n(n-1)/2

Using this result for the arithmetic series above, we find:

s_n = n a₁ + r n(n - 1)/2
We have:
a_n = (n - 1)r + a₁
so
s_n = n a₁ + r n(n - 1)/2 = (n/2)[ 2a₁ + r (n - 1)] =
(n/2)[ a₁ + a₁ + r (n - 1)] =
(n/2)[ a₁ + a_n]

a_n = (n - 1)r + a₁

s_n = n a₁ + r n(n - 1)/2 =
(n/2)[ a₁ + a_n]

To determine whether a series is geometric, we evaluate the difference a_n+1 - a_n. If this ratio is constant, the series is arithmetic.

lim s_n = + ∞. The series is divergent.
n → + ∞

2. Example

The sequence {a_n} = {1, 4, 7, 10, ...} has the general term a_n = 3(n - 1) + 1

The sequence is arithmetic. Indeed:

a_{n + 1} - a_n = 3n + 1 - 3(n - 1) - 1 = 3 .
This difference is constant.

The common difference equal to 3, and the first tern a₁ is 1. We can verify then that the general term is:
a_n = 3(n - 1) + 1

The related series has the general term:
s_n = n a₁ + r n(n - 1)/2 = (n/2)[ a₁ + a_n] =
n .1 + 3 n(n - 1)/2 = n(3n - 1)/2