Σ(1 / (a^n)) from n=1 to m
This blog entry covers the sum of the series:
Σ[1 / (a^n), n=1 to m] with n and m positive integers
Specific Cases: a = 2 and a = 3
When a = 2:
m = 1: 1/2
m = 2: 1/2 + 1/4 = (2 + 1)/4 = 3/4
m = 3: 1/2 + 1/4 + 1/8 = (4 + 2 + 1)/8 = 7/8
m = 4: 1/2 + 1/4 + 1/8 + 1/16 = (8 + 4 + 2 + 1)/16 = 15/16
Going from the pattern,
Σ[1 / (2^n), n=1 to m] = 1/(2^m) * Σ[(2^n), n=0 to m-1] = (2^m - 1) / 2^m
When a = 3:
m = 1: 1/3
m = 2: 1/3 + 1/9 = (3 + 1)/9 = 4/9
m = 3: 1/3 + 1/9 + 1/27 = (9 + 3 + 1)/27 = 13/27
m = 4: 1/3 + 1/9 + 1/27 + 1/81 = (27 + 9 + 3 + 1)/81 = 40/81
Going from the pattern,
Σ[1 / (3^n), n=1 to m] = 1/(3^m) * Σ[(3^n), n=0 to m-1]
Finding the General Formula and Proof
Let's presume that, for any a:
Σ[1 / (a^n), n=1 to m-1] = 1/(a^(m-1) * Σ[(a^n), n=0 to m-2]
Let's add 1/(a^m) to the series:
Σ[1 / (a^n), n=1 to m-1] + 1/(a^m)
= (1 + a + a^2 + ... + a^(m-3) + a^(m-2)) / (a^(m-1)) + 1 / (a^m)
= (a * (1 + a + a^2 + ... + a^(m-3) + a^(m-2)) + 1) / (a^m)
= (a + a^2 + a^3 + ... + a^(m-2) + a^(m-1) + 1) / (a^m)
= 1/(a^(m)) * Σ[(a^n), n=0 to m-1]
The general formula is now:
Σ[1 / (a^n), n=1 to m] = 1/(a^(m)) * Σ[(a^n), n=0 to m-1]
Until next time,
Eddie
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