Not sure about this one:
Find the sum of the infinite geometric series 1+3/5+9/25+27/125+...
Thanks for any help.
In geometric sequence there is a common ratio, r, between two successive terms.
a2 = a1 *r
a3 = a2 *r = a1 *r^2
....
an = a1 *r^(n-1)
In a geometric series up to infinity, if the absolute value of r is less than 1, then the series converges and its sum is a1/(1-r).
1 +3/5 +9/25 +27/125 +...
r = (3/5)/1 = 3/5
r = (9/25)/(3/5) = (9/25)*(5/3) = 3/5
r = (27/125)/(9/25) = (27/125)*(25/9) = 3/5
Ok, r = 3/5 = 0.6 ----less than 1, so there is a sum.
Sum = a1 / (1-r)
Sum = 1 / (1 -3/5) = 1/(2/5) = 5/2 -------------answer.
Hello, Mike!
There is a formula for the sum of an infinite geometric series.
. . . . . . . . a
. . S . = . ------
. . . . . . . 1 - r
where a = first term, r = common ratio.
Find the sum of the infinite geometric series: 1 + 3/5 + 9/25 + 27/125 + ...
We have: a = 1, r = 3/5
. . . . . . . . . . . .1 . . . . . . 1 . . . . 5
Then: . S .= .--------- .= .----- .= . --
. . . . . . . . . .1 - 3/5 . . . 2/5 . - - -2