Evaluate
lim 2r Σ r=1 [2n^2/(n+r)^3]
n→∞
limn→∞∑r=12n2n2(n+r)3=limn→∞∑r=12n2n2n3(1+rn)3=limn→∞1n∑r=12n2(1+rn)3=\lim\limits_{n\rightarrow\infty}{\sum_{r=1}^{2n}\frac{2n^2}{\left(n+r\right)^3}}=\lim\limits_{n\rightarrow\infty}{\sum_{r=1}^{2n}\frac{2n^2}{{n^3\left(1+\frac{r}{n}\right)}^3}}=\lim\limits_{n\rightarrow\infty}{\frac{1}{n}\sum_{r=1}^{2n}\frac{2}{\left(1+\frac{r}{n}\right)^3}}=n→∞lim∑r=12n(n+r)32n2=n→∞lim∑r=12nn3(1+nr)32n2=n→∞limn1∑r=12n(1+nr)32=
=2∫limn→∞1n2dx(1+x)3=2∫02d(1+x)(1+x)3==2\int_{\lim\limits_{n\rightarrow\infty}{\frac{1}{n}}}^{2}\frac{dx}{\left(1+x\right)^3}=2\int_{0}^{2}\frac{d\left(1+x\right)}{\left(1+x\right)^3}==2∫n→∞limn12(1+x)3dx=2∫02(1+x)3d(1+x)=
=2(1+x)−2−2∣20=−1(1+x)2∣20=−19+1=89=2\left.\frac{\left(1+x\right)^{-2}}{-2}\right|\begin{matrix}2\\0\\\end{matrix}=\left.-\frac{1}{\left(1+x\right)^2}\right|\begin{matrix}2\\0\\\end{matrix}=-\frac{1}{9}+1=\frac{8}{9}=2−2(1+x)−2∣∣20=−(1+x)21∣∣20=−91+1=98
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