Question

Evaluate

lim 3nΣr=1 n^2/(4n+r)^3

n→∞

Answer 1

By integral test for series,

$lim_{n\to\infty} \Sigma^{3n}_1 \frac{n^2}{(4n+x)^3}\\ =lim_{n\to\infty} \Sigma^{3n}_1 \frac{1}{n(4+x)^3}\\ =\int^3_1 (x+4)^{-3}dx\\ =\frac{1}{-2}[(x+4)^{-2}]_1^3\\ =\frac{1}{-2}[\frac{1}{49}-\frac{1}{25}]\\ =\frac{2}{1225}$