The series is convergent
from x=? to x=?
1)
∞
∑ [(X/X6)^n]/n(6)^n
n=1
2)
∞
∑ [(11x)^n]/n^10
n=1
3)
∞
∑ [(x9)^n]/9^n
n=1
4)
∞
∑ [(1)^n*x^n]/2^n(n^2+6)
n=1
from x=? to x=?
1)
∞
∑ [(X/X6)^n]/n(6)^n
n=1
2)
∞
∑ [(11x)^n]/n^10
n=1
3)
∞
∑ [(x9)^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].

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 pseries
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].

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).

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 pseries (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].

I hope this helps!
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].

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 pseries
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].

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).

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 pseries (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].

I hope this helps!