1. ## Sequence Series Question

Find a geometric series with common ration 1/w that has limiting sum 1/(1-w)

2. Originally Posted by acevipa
Find a geometric series with common ration 1/w that has limiting sum 1/(1-w)
Solve $\displaystyle \frac{1}{1-w} = \frac{a}{1 - \frac{1}{w}}$ for $\displaystyle w$ subject to the restriction $\displaystyle \left| \frac{1}{w}\right| < 1$ (a is the first term in the series).

3. Originally Posted by acevipa
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