Show that (i) limn → ∞ (1 + 2 + 3 + ... + n)/(3n^2 + 7n + 2) = 1/6 (ii) limn → ∞ (1^2 + 2^2 + ... + (3n)^2)/((1 + 2 + ... + 5n)(2n + 3)) = 9/25

Question

Show that 

(i) \( \lim\limits_{n \to ∞} {\frac{1 + 2 + 3 + ... + n }{3n^2 + 7n + 2}}\) = 1/6

(ii) \( \lim\limits_{n \to ∞} {\frac{1^2 + 2^2 + ... + (3n)^2 }{(1 + 2 + ... + 5n) (2n + 3) }}\) = 9/25

(iii) \( \lim\limits_{n \to ∞} {\frac{1}{1.2}} + {\frac{1}{2.3}} + {\frac{1}{3.4}} +... + {\frac{1}{n( n + 1)}}\) = 1

Answer 1

(i) \( \lim\limits_{n \to ∞} {\frac{1 + 2 + 3 + ... + n }{3n^2 + 7n + 2}}\)

(ii) \( \lim\limits_{n \to ∞} {\frac{1^2 + 2^2 + ... + (3n)^2 }{(1 + 2 + ... + 5n) (2n + 3) }}\) 

(iii) \( \lim\limits_{n \to ∞} {\frac{1}{1.2}} + {\frac{1}{2.3}} + {\frac{1}{3.4}} +... + {\frac{1}{n( n + 1)}}\)