Question

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.

Answer 1

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)