Σ(1 / (a^n)) from n=1 to m

 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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