Find the interval of convergence for the given power series:
Sum[(x-5)^n / n(-3)^n] n=1 to infinity
My attempt thus far:
I think I need to do the ratio test?
lim[(x-5)^(n+1) / (n+1)(-3)^(n+1) * n(-3)^n / (x-5)^n] of n to infinity
simplify:
lim[(x-5)n / ((n+1)(-3)^(n+1))]
|x-5|*lim[n / -3(n+1)]
Do I need to try and move that -3 to outside the limit? How do I do that?
Then I tested the endpoints and got that they both converged, but that's assuming I did first part correctly, which I evidently didn't. Could someone please explain this to me? Thanks!
Sum[(x-5)^n / n(-3)^n] n=1 to infinity
My attempt thus far:
I think I need to do the ratio test?
lim[(x-5)^(n+1) / (n+1)(-3)^(n+1) * n(-3)^n / (x-5)^n] of n to infinity
simplify:
lim[(x-5)n / ((n+1)(-3)^(n+1))]
|x-5|*lim[n / -3(n+1)]
Do I need to try and move that -3 to outside the limit? How do I do that?
Then I tested the endpoints and got that they both converged, but that's assuming I did first part correctly, which I evidently didn't. Could someone please explain this to me? Thanks!
-
1st : Ratio Test uses | a_(n+1) / a_n |...thus the ' - 3 ' is really + 3
2nd : lim {n--->∞ } = | x - 5} / 3 < 1---> ( 2 , 8 )..at x = 8 converges , x = 2 , diverges
2nd : lim {n--->∞ } = | x - 5} / 3 < 1---> ( 2 , 8 )..at x = 8 converges , x = 2 , diverges