1. The radius of convergence is ∞.

1. Factorials suggest the Ratio test. lim_{n → ∞}|((n+1)!xⁿ⁺¹/(2n+2)!) / (n!x_n/(2n)!)| = lim_{n → ∞}|((n+1)!/n!)⋅((2n)!/(2n+2)!)x| = lim_{n → ∞}(n+1)/((2n+2)(2n+1))|x| = 0 for all x. This power series converges for all x, so the radius of convergence is ∞.

1. The radius of convergence is 0. The interval of convergence is [0].

1. Exponents suggest the Root Test. lim_{n → ∞}(nⁿ|x|ⁿ/2ⁿ)^1/n= lim_{n → ∞}(n/2)|x| = ∞ for all x ≠ 0. Then the radius of convergence is 0, and the inverval of convergence is [0].

1. The radius of convergence is 2. The interval of convergence is (-2, 2).

1. The exponent of n² suggests the Root Test will not be effective. Try the Ratio Test. lim_{n → ∞}|((n+1)²xⁿ⁺¹/2ⁿ⁺¹) / (n²xⁿ/2ⁿ)| = lim_{n → ∞}((n+1)²/n²)⋅(2ⁿ/2ⁿ⁺¹)|x| = |x|/2. The interval of convergence includes |x|/2 < 1, that is, (-2,2), and so the radius of convergence is 2. To find the interval of convergence, test the endpoints of (-2,2). For x = 2 the series becomes ∑_n=1^∞ n²2ⁿ/2ⁿ = ∑_n=1^∞ n². This diverges by the n^th term test. For x = -2 the series becomes ∑_n=1^∞ n²(-2)ⁿ/2ⁿ = ∑_n=1^∞ (-1)ⁿn². This diverges by the n^th term test. Then the interval of convergence is (-2,2).

1. The radius of convergence is 1/2. The interval of convergence is (0, 1).

1. The exponent of n^1/2 suggests the Root Test will not be effective. Try the Ratio Test. lim_{n → ∞}|((2x – 1)²ⁿ⁺³/(n+1)^1/2) / ((2x – 1)²ⁿ⁺¹/n^1/2)| = lim_{n → ∞}(n/(n+1))^1/2|(2x – 1)²| = |(2x – 1)²| The interval of convergence includes |(2x – 1)²| < 1, that is, |2x – 1| < 1, which gives 0 < x < 1. Then the radius of convergence is 1/2. To find the interval of convergence, test the endpoints of (0,1). For x = 1 the series becomes ∑_n=1^∞ 1²ⁿ⁺¹/n^1/2 = ∑_n=1^∞ 1/n^1/2, a diverging p-series. For x = 0 the series becomes ∑_n=1^∞ (-1)²ⁿ⁺¹/n^1/2 = -∑_n=1^∞ 1/n^1/2, a diverging p-series. Then the interval of convergence is (0,1).