Find the interval of convergence for the power series~~~~~~~~~~~~~~~~~~help

The series is convergent
from x=? to x=?


1)

∞
∑ [(X/X-6)^n]/n(-6)^n
n=1


2)

∞
∑ [(11x)^n]/n^10
n=1


3)

∞
∑ [(x-9)^n]/9^n
n=1


4)
∞
∑ [(-1)^n*x^n]/2^n(n^2+6)
n=1

1) Is this series actually a power series?

Assuming you meant (x - 6)^n:

Using the ratio test:
r = lim(n→∞) | [(x - 6)^(n+1) / ((n+1) (-6)^(n+1))] / [(x - 6)^n / (n(-6)^n)] |
..= (1/6) |x - 6| * lim(n→∞) n/(n+1)
..= (1/6)|x - 6|.

So, the series converges (at least) for (1/6) |x - 6| < 1
==> |x - 6| < 6.

Checking the endpoints:
x = 0 ==> ∑(n = 1 to ∞) 1/n, divergent harmonic series
x = 12 ==> ∑(n = 1 to ∞) (-1)^n /n, convergent by alternating series test.

So, the interval of convergence is (0, 12].
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2) Using the ratio test:
r = lim(n→∞) | [(11x)^(n+1) / (n+1)^10] / [(11x)^n / n^10] |
..= 11 |x| * lim(n→∞) (n/(n+1))^10
..= 11 |x|.

So, the series converges (at least) for 11 |x| < 1 ==> |x| < 1/11.

Checking the endpoints:
x = 1/11 ==> ∑(n = 1 to ∞) 1/n^10, convergent p-series
x = -1/11 ==> ∑(n = 1 to ∞) (-1)^n /n^10, converges absolutely (by previous remark)

So, the interval of convergence is [-1/11, 1/11].
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3) This is a geometric series with r = (x - 9)/9.
So, the series converges for |x - 9| / 9 < 1 ==> x is in (0, 18).
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4) Using the ratio test:
r = lim(n→∞) | [(-1)^(n+1) x^(n+1) / (2^(n+1) ((n+1)^2 + 6))] / [(-1)^n x^n / (2^n (n^2 + 6))] |
..= (1/2) |x| * lim(n→∞) (n^2 + 6) / ((n+1)^2 + 6)
..= (1/2) |x|.

So, the series converges (at least) for |x|/2 < 1 ==> |x| < 2.

Checking the endpoints:
x = -2 ==> ∑(n = 1 to ∞) 1/(n^2 + 6)), convergent upon comparison with p-series (p = 2)
x = 2 ==> ∑(n = 1 to ∞) (-1)^n / (n^2 + 6), converges absolutely (by previous remark)

So, the interval of convergence is [-2, 2].
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I hope this helps!