Question 1092428: How do I determine whether the sequence converge or diverges. If it converges, give the limit.
U1=1, U n+1= Un/3 for n≥1 (1,n+1, n after U should be smaller size).
Please help. Thank you.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
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.
http://mathematics.laerd.com/maths/geometric-progression-intro.php
|
|
|
