lim n → ∞ ∑ r = 1 2 n 2 n 2 ( n + r ) 3 = lim n → ∞ ∑ r = 1 2 n 2 n 2 n 3 ( 1 + r n ) 3 = lim n → ∞ 1 n ∑ r = 1 2 n 2 ( 1 + r n ) 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 = 1 2 n ( n + r ) 3 2 n 2 = n → ∞ lim ∑ r = 1 2 n n 3 ( 1 + n r ) 3 2 n 2 = n → ∞ lim n 1 ∑ r = 1 2 n ( 1 + n r ) 3 2 =
= 2 ∫ lim n → ∞ 1 n 2 d x ( 1 + x ) 3 = 2 ∫ 0 2 d ( 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 → ∞ l i m n 1 2 ( 1 + x ) 3 d x = 2 ∫ 0 2 ( 1 + x ) 3 d ( 1 + x ) =
= 2 ( 1 + x ) − 2 − 2 ∣ 2 0 = − 1 ( 1 + x ) 2 ∣ 2 0 = − 1 9 + 1 = 8 9 =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 ∣ ∣ 2 0 = − ( 1 + x ) 2 1 ∣ ∣ 2 0 = − 9 1 + 1 = 9 8
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