Thread: Convergence question (Calc)

For the series below, I have to prove that it either converges, diverges, or that whether it converges/diverges cannot be determined. I'm not sure how to go about this, so any help would be appreciated!

Sigma starting at j=1 and ending at infinity of (2/3)^j + (3/4)^j

I mean, I know each of the individual terms in the series converge because they are geometric. (2/3)^j converges to 3 and (3/4)^j converges to 4. Does this mean I can just add them to get 7 and because of this, it converges?

Originally Posted by clockingly

For the series below, I have to prove that it either converges, diverges, or that whether it converges/diverges cannot be determined. I'm not sure how to go about this, so any help would be appreciated!

Sigma starting at j=1 and ending at infinity of (2/3)^j + (3/4)^j

I mean, I know each of the individual terms in the series converge because they are geometric. (2/3)^j converges to 3 and (3/4)^j converges to 4. Does this mean I can just add them to get 7 and because of this, it converges?

Hello,

the proposed way to do the problem is OK, because both summands have the same exponent - but since

∑[from 1 to ∞](2/3)^j = 2 and

∑[from 1 to ∞](3/4)^j = 3

the series converges against 5 (or do you say to 5 (?)).