Evaluate
lim 3nΣr=1 n^2/(4n+r)^3
n→∞
By integral test for series,
limn→∞Σ13nn2(4n+x)3=limn→∞Σ13n1n(4+x)3=∫13(x+4)−3dx=1−2[(x+4)−2]13=1−2[149−125]=21225lim_{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}limn→∞Σ13n(4n+x)3n2=limn→∞Σ13nn(4+x)31=∫13(x+4)−3dx=−21[(x+4)−2]13=−21[491−251]=12252
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