I have been given the following info:
a<1> = 1, a<n+1> = 3*a<n>/6+a<n>
I plugged L in and ended up with the equation: L = 3*L/6+L, which I solved and got L = -3 for the limit. However, this answer is either only partially correct or close to the true answer. Any help is appreciated.
A little work shows that the nth term
a<n> = 3/(2^(n+1)-1)
Now, do you want the limit of the terms, or the limit of the sums?
a<?> = 0
S<?> = 3(sum 1/(2^(n+1)-1)
I thought that would be relatively simple, but you can see below that it is not:
http://www.wolframalpha.com/input/?i=sum+1%2F(2%5Ek-1)
Based on the equation you provided, we have the recurrence relation:
a<1> = 1
a<n+1> = 3 * a<n> / 6 + a<n>
To find the limit (if it exists) of this sequence, we can express it in terms of itself:
L = 3 * L / 6 + L
Simplifying this equation, we get:
L = (3L + 6L) / 6
L = 9L / 6
To solve for L, we can multiply both sides of the equation by 6:
6L = 9L
Subtracting 6L on both sides:
6L - 6L = 9L - 6L
0 = 3L
Dividing by 3:
L = 0
Therefore, the limit of the given sequence is L = 0.