If |r| > 1 then the sum of the series is infinity, so presumable it means not.
Also are you sure about the second series: 1-r+r2 the 'minus' ?
Last edited by TrueGrit; 22-06-2012 at 21:00.
So, it took you 13 hours to get around to reading the question properly.
If |r|<1 then as you add terms the sum approaches some particular value. For example S = 1+1/2+1/4+1/8+1/16......
The sums are
and it's clearly approaching 2 and you can get as close to 2 as you like.
Last edited by BabyMaths; 23-06-2012 at 08:43.
For infinite geometric series
(Original post by Julii92)
The sum of the geometric series 1+r+r2
+... is k
times the sum of the series 1-r+r2
..., where k>0. Express r
in terms of k
By equating the sums of both series, I've got (1-rn
)/(1-r) = k(1-(-r)n
)/(1+r), but I'm struggling from here. Help appreciated.
So your equation
Solve for r
then from |r|<1 determine which integer can be k (maybe it's only my question)
Last edited by ztibor; 24-06-2012 at 09:36.