1. Converges to 1/2.

2. Factoring n² from numerator and denominator, we find limn_{→ ∞} (n² + 3)/(2n² – 5) = lim_{n<→ ∞}(1 + 3/n²)/(2 – 5/n²) = (1 + 0)/(2 – 0) = 1/2

1. Converges to 0.

1. Factoring n³ from numerator and denominator, we find lim_{n → ∞}(n² + 3)/(2n³ – 5)= lim_{n → ∞}(1/n + 3/n³)/(2 – 5/n³) = (0 + 0)/(2 – 0) = 0/2 = 0

1. Diverges

1. Factoring n from numerator and denominator, we find lim_{n → ∞} (n² + 3)/(2n – 5)= lim_{n → ∞}(n + 3/n)/(2 – 5/n) =lim_{n → ∞} n/2= ∞

1. Converges to 0

1. Multiply by (√(n+1) + √n)/(√(n+1) + √n), obtaining lim_{n → ∞} √(n+1) – √n = lim_{n → ∞} (√(n+1) – √n)(√(n+1) + √n)/(√(n+1) + √n) = lim_{n → ∞} (n + 1 – n)/(√(n+1) + √n) = lim_{n → ∞} 1/(√(n+1) + √n) = 0

1. Diverges

1. Because lim_{n → ∞} n²/(2n² + 1)= lim_{n → ∞} 1/(2 + 1/n²) = 1/(2 + 0) = 1/2 the series diverges by the 9^th term test.

1. Converges to 3/2

1. Apply the method of partial fractions and conclude: 2/(n² – 1) = 1/(n – 1) – 1/(n + 1) Starting the series from n = 2, the first few terms are (1/(2-1) – 1/(2+1)) + (1/(3-1) – 1/(3+1)) + (1/(4-1) – 1/(4+1)) + (1/(5-1) – 1/(5+1)) + … = (1 – 1/3) + (1/2 – 1/4) + (1/3 – 1/5) + (1/4 – 1/6) + … = 1 + 1/2 = 3/2 because 1/3 occurs twice, once + and once -, 1/4 occurs twice, once + and once -, and so on. Only the 1 and 1/2 are not cancelled by later terms.

1. Diverges

1. Note that n sin(1/n) = sin(1/n)/(1/n). Because 1/n → 0 as n → ∞, and recalling lim_{x → 0} sin(x)/x = 1, we see lim_{n → ∞}n sin(1/n) = 1. The series diverges by the n^th term test.

1. Diverges

1. As n → ∞, arctan(n) → π/2, so the series diverges by the n^th term test.