Find a geometric series with common ration 1/w that has limiting sum 1/(1-w)
hey mate,
good question
As Im sure you well aware a geometric series a(n) is defined so that
a(n+1) = r * a(n), where r is the common ratio
here you have given that r = 1/w
thus,
a(n+1) = (1/w)a(n)
assuming a(0) = a you then have a(n) = (1/w)^n * a(0) = a * (1/w)^n
furthermore you have stated the the limiting sum convergences to 1/(1-w), i.e.
sum(a(n), n = 0..inf) = 1/(1-w)
or
sum(a * (1/w)^n, n = 0..inf) = 1/(1-w)
since a is a constant, bring it out the front of the sum
thus
= a * sum ( (1/w)^n, n = 0...inf) = 1/(1-w)
The sum component is known as the sum of a geometric series, whereby
a(n) is geometric, then
sum(a(n), n = 0...inf)) = 1/(1-r)
where r is the common ratio, here r = 1/w
Thus the 'sum' component of the previous equation can be replaced with
1/ (1- (1/w)) = w/(w-1)
hence we obtain
a * w/(w-1) = 1/(w-1)
hence a = 1/w
Thus, a(n) = a * (1/w)^n = (1/w)^(n+1), n = 0... inf
Hope this helps,
Let me know if you require any further clarrification and if I have made any mistakes,
Regards,
David