is this a convergent sequence, if so what is its limit?

Question

(n+2(3^n)+3(2^n)) / (n^2-5(2^n)+4(3^n)) for n+1,2,.....

Answer 1

If you want a quick answer with a minimal amount of work, here's something to look for.
What are the dominant terms as n gets larger? They will be the terms involving 3^n.
So as n gets larger, the expression becomes more and more like:
2*(3^n) / 4*(3^n) = 1/2

NOTE:
This means that the SEQUENCE converges. However, the SUMMATION of the series does not converge because the last term is not zero.

Answer 2

Since the numerator and denominator are dominated
by the 3^n terms, the sequence converges to the
ratio of their coefficients, i.e. to 2/4 which equals 1/2.

Answer 3

(n+2(3^n)+3(2^n)) / (n^2-5(2^n)+4(3^n)) will be dominated by the 3^n terms as n gets real large. The limit as n goes to infinity would be the same as the limit of 2(3^n)/4(3^n) goes to infinity, which is just 2/4 = 1/2.

Answer 4

yes converges to 0.5.
as n goes to infinity, the most important term
in the numerator is 2(3^n) and that in the denominator is 4(3^n)
the ratio goes to half.
or if you want, divide the numerator and denominator by the function 3^n. you will see that you get 2 + n/3^n + 3(2^n/3^n) in the numerator, which approaches 2.
similarly the denominator approaches 4.

Answer 5

Yes. As n goes to infinity, the value of the function converges to 1/2.