lim n → ∞ n 2 n 2 + 1 n + 1 + n 2 n 2 + 4 n + 2 + n 2 n 2 + 9 n + 3 + … + n 2 n 2 + n 2 n + n is equal to

Step-by-step solution

lim n → ∞ ∑ r = 1 n 1 n 1 + r 2 n 2 1 + r n = ∫ 0 1 d x 1 + x 2 1 + x Let I = ∫ 0 1 d x 1 + x 2 1 + x Now x = tan θ , d x = sec 2 θ d θ I = ∫ 0 π 4 d θ 1 + tan θ = 1 2 ∫ 0 π 4 2 cos θ d θ sin θ + cos θ = 1 2 ∫ 0 π 4 sin θ + cos θ d θ sin θ + cos θ + 1 2 ∫ 0 π 4 cos θ − sin θ d θ sin θ + cos θ = 1 2 × π 4 + 1 2 ln sin θ + cos &#9