SOLUTION: 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

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

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SOLUTION: 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&#8805;1 (1,n+1, n after U should be smaller size). Please

SOLUTION: 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