Evaluate
lim 3nΣr=1 n^2/(4n+r)^3
n→∞
By integral test for series,
"lim_{n\\to\\infty} \\Sigma^{3n}_1 \\frac{n^2}{(4n+x)^3}\\\\\n=lim_{n\\to\\infty} \\Sigma^{3n}_1 \\frac{1}{n(4+x)^3}\\\\\n=\\int^3_1 (x+4)^{-3}dx\\\\\n=\\frac{1}{-2}[(x+4)^{-2}]_1^3\\\\\n=\\frac{1}{-2}[\\frac{1}{49}-\\frac{1}{25}]\\\\\n=\\frac{2}{1225}"
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