Question 1092428


if abs(r) < 1, then the geometric sum will converge.


if abs(r) > 1, then the geometric sum will not converge.


abs(r) < 1 means that:


r < 1 or -r < 1


solve for r in each of these equations and you get:


r < 1 is already solved for r.


start with -r < 1


multiply both sides of this equation by -1 to get:


r > -1


you get abs(r) < 1 when r < 1 and when r > -1.


this means -1 < r < 1


the geometric series will converge when that occurs.


since your common ratio is 1/3 which is > -1 and < 1, then your geometric series will converge.


the formula for the sum of a geometric series is:


Sn = A1 * (1 - r^n) / (1-r)


when -1 < r < 1, this series will converge to a limit as n approaches infinity.


that's because r^n will get closer and closer to 0 as n approaches infinity.


the limit of Sn as n approaches infinity when -1 < r < 1 is:


limit of Sn as n approaches infinity = 1 / (1-r).


since r = 1/3 in your problem, you get:


limit of Sn as n approaches infinity = 1 / (1-(1/3)).


this results in limit of Sn as n approaches infinity = 3/2 or 1.5.


here's a reference on the sum of a geometric series.


<a href = "http://mathematics.laerd.com/maths/geometric-progression-intro.php" target = "_blank">http://mathematics.laerd.com/maths/geometric-progression-intro.php</a>