Um, Intervals of convergance. (advanced junk)?

Question

I'm sposed to find the interval of convergence for the power series defined by (-1)^n (n(x^n))/(n+1)

I did a ratio test on it and I came up with -1

Answer 1

If you've done the ratio test, then you've already justified it. Well, you do have to test the endpoints separately (since a limiting ratio of 1 is technically inconclusive), but this is not hard, as both x=1 and x=-1 fail the nth term test. Therefore, the series converges on the open interval (-1, 1).

Answer 2

To test -1 < x < 1:

x^n converges, and n/(n+1) * x^n < x^n for all n > 0. Therefore, n/(n+1) * x^n must also converge.

To test x = 1:

The function becomes ±n/(n+1), which approaches ±1 as n gets large. So, it diverges.

To test x > 1 or x < -1:

As noted above, n/(n+1) approaches ±1 as n gets large. x^n also does not approach zero if x >1 or x < -1. Since neither part approaches zero, the product cannot converge.

Answer 3

take pairs of terms

p[n] = 2n x^(2n) / (2n+1) - (2n+1)x^(2n+1) / (2n+2)
= x^(2n) / [(2n+1)(2n+2) ] [ 2n(2n+2) - (2n+1)(2n+1)x ]
= x^(2n) / [(2n+1)(2n+2)] [ (2n+1 - 1)(2n+2) -(2n+1)(2n+2-1)x ]
= x^(2n) / [(2n+1)(2n+2)] [ x(2n+1)-(2n+2)]
= x^(2n) / [(2n+1)(2n+2)] [ (x-1)(2n+1)-1]
if |x| <= 1 this converges