1. The radius of convergence is 1, the interval of convergence is [-1,1).

1. Apply the Ratio Test lim_{n → ∞}|(xn+1/√(n+1))/(xⁿ/√n)| = lim_{n → ∞} |x|⋅√(n/(n+1)) = |x| By the Ratio Test, the series converges for |x| < 1, that is, -1 < x < 1. The radius of convergence is 1. At the left endpoint, the series becomes ∑_n=1^∞(-1)ⁿ/√n. This converges by the Alternating Series Test. At the right endpoint, the series becomes ∑_n=1^∞1/√n. This is a p-series with p = 1/2, so diverges. The interval of convergence is [-1,1).

1. The radius of convergence is 5, the interval of convergence is [-4,6).

1. Apply the Ratio Test lim_{n → ∞}|((x-1)ⁿ⁺¹/((n+1)5ⁿ⁺¹) / ((x-1)ⁿ/(n 5ⁿ)| = lim_{n → ∞} |(x-1)ⁿ⁺¹/(x-1)ⁿ|⋅(n/(n+1))⋅(5ⁿ/5ⁿ⁺¹) = |x-1|/5 By the Ratio Test, the series converges for |x-1|/5 < 1, that is for -5 < x-1 < 5, or -4 < x < 6. The radius of convergence is 5. At the left endpoint the series becomes ∑_n=1^∞(-5)ⁿ/(n 5ⁿ) = ∑_n=1^∞(-1)ⁿ/n convergent by the Alternating Series Test. At the right endpoint the series becomes ∑_n=1^∞(5)ⁿ/(n 5ⁿ)= ∑_n=1^∞ 1/n This is the harmonic series, hence divergent. The interval of convergence is [-4, 6).

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

1. Apply the Ratio Test lim_{n → ∞}|((-1)ⁿ⁺¹xⁿ⁺¹)/(2ⁿ⁺¹ln(n+1)) / ((-1)ⁿxⁿ)/2nln(n))| = lim_{n → ∞} |-(xn+1/xn)|⋅(2n/2n+1)⋅(ln(n)/ln(n+1)) = |-x|/2 where limn → ∞ln(n)/ln(n+1) = 1 is obtained by applying l’Hôpital’s rule to lim_{x → ∞}ln(x)/ln(x+1) By the Ratio Test, the series converges for |-x|/2 < 1, that is, for -2 < x < 2. The radius of convergence is 2. At the left endpoint, the series becomes ∑_n=2^∞(-1)ⁿ(-2)ⁿ/(2ⁿln(n))= ∑_n=2^∞ 2ⁿ/(2ⁿ ln(n))) = ∑_n=2^∞1/ln(n) Now ln(n) < n, so 1/ln(n) > 1/n and the series diverges by comparison with the harmonic series. At the right endpoint, the series becomes ∑n=2∞(-1)n2n/(2nln(n)) = ∑_n=2^∞ (-1)ⁿ/ln(n) converging by the Alternating Series Test. The interval of convergence is (-2,2].

1. The radius of convergence is 1/3, the interval of convergence is [-2/3, 0].

1. Apply the Ratio Test lim_{n → ∞}|((3x+1)ⁿ⁺¹/(n+1)²) / ((3x+1)ⁿ/n²)|= lim_{n → ∞} |(3x+1)ⁿ⁺¹/(3x+1)ⁿ|⋅(n²/(n+1)²)= |3x+1| By the Ratio Test, the series converges for |3x+1| < 1, that is, -1 < 3x+1 < 1, so -2 < 3x < 0, giving -2/3 < x < 0. The radius of convergence is 1/3. At the left endpoint, the series becomes ∑_n=1^∞ (-1)ⁿ/n² convergent by the Alternating Series Test. At the right endpoint, the series becomes ∑_n=1^∞ 1ⁿ/n² convergent, being a p-series with p= 2. The interval of convergence is [-2/3, 0].

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

1. Apply the Ratio Test: lim_{n → ∞}|(xⁿ⁺¹/(n+1)ⁿ⁺¹)/(xⁿ/nⁿ)| = lim_{n → ∞} |xⁿ⁺¹/xⁿ|⋅((nⁿ)/((n+1)ⁿ⁺¹)) = lim_{n → ∞}|x|⋅(n/(n+1))ⁿ⋅(1/(n+1)) = 0 for all x Here we have used lim_{n → ∞}(1 + 1/n)ⁿ= e, we see lim_{n → ∞}(n/(n+1))ⁿ= 1/e to be sure we needn’t be concerned with the factor (n/(n+1))ⁿ in the limit. By the Ratio Test, this series converges for al x, so the radius of convergence is ∞ and the interval of convergence is (-∞, ∞).