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# help

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I don't where to start on this problem:

Compute

$$\left( \frac{1}{2} \right)^1 + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^4 + \left( \frac{1}{2} \right)^5 + \left( \frac{1}{2} \right)^7 + \left( \frac{1}{2} \right)^8 + \dotsb$$

where the exponents are all positive integers, except multiples of 3.

May 11, 2020

#1
+1

We  can  evalute  this  series

(1/2) +  (1/2)^2  + (1/2)^3  + (1/2)^4   + (1/5)^5  + (1/2)^6 + (1/2)^7  + (1/2)*8  +   .......

The   sum of this geometric  series    =       (a) / (  1 - r)   where a  is the first term  and r is  the common ratio

The sum is    (1/2)  / (1 - 1/2)   =   (1/2)  / (1/2)  =   1

And  from this sum we can subtract this sum

(1/2)^3  + ( 1/2)^6  +  (1/2)^9  + (1/12)^12  +  ..........

The first term  = (1/2)^3    and the common ratio  = (1/2)^3

( 1/2)^3  /  [  1  - (1/2)^3]   =    (1/8)  /  [ 1 - 1/8]  =  (1/8) / (7/8)  =  1/7

So....the  given series  evaluates to

1 -  1/7   =

6/7   May 11, 2020
edited by CPhill  May 11, 2020