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      Calculus II

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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
... = ...
an = (n - 1)r + a1

The sum of the first terms is:
sn = a1 + a2 + ... + an
= 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:

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

an = (n - 1)r + a1

sn = n a1 + r n(n - 1)/2 =
(n/2)[ a1 + an]


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

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


2. Example

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

The sequence is arithmetic. Indeed:

an + 1 - an = 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:
an = 3(n - 1) + 1

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



sn = 1 + 4 + 7 + 10 + ... + 3(k - 1) + 1 = n(3n - 1)/2
k: 1 → n



  


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