Answer to Question #278661 in Real Analysis for Reck

Question #278661

Evaluate


lim 3nΣr=1 n^2/(4n+r)^3


n→∞



1
Expert's answer
2021-12-14T15:01:04-0500

By integral test for series,

limnΣ13nn2(4n+x)3=limnΣ13n1n(4+x)3=13(x+4)3dx=12[(x+4)2]13=12[149125]=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}


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