Sunday, October 17, 2021

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


All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 


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