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$\lim _{n \rightarrow \infty}\left(\frac{1^2}{n^3+1^3}+\frac{2^2}{n^3+2^3}+\ldots+\frac{1}{2 n}\right)=$
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The correct answer is:
$\log \sqrt[3]{2}$
$L=\lim _{n \rightarrow \infty}\left[\frac{1^2}{n^3+1^3}+\frac{2^2}{n^3+2^3}+\ldots+\frac{n^2}{n^3+n^3}\right]$
$=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3+r^3}=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3\left(1+\frac{r^3}{n^3}\right)}$
$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{\left(\frac{r}{n}\right)^2}{1+\left(\frac{r}{n}\right)^3}=\int_0^1 \frac{x^2 d x}{1+x^3}$
$=\frac{1}{3} \log 2=\log \sqrt[3]{2}$
$=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3+r^3}=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3\left(1+\frac{r^3}{n^3}\right)}$
$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{\left(\frac{r}{n}\right)^2}{1+\left(\frac{r}{n}\right)^3}=\int_0^1 \frac{x^2 d x}{1+x^3}$
$=\frac{1}{3} \log 2=\log \sqrt[3]{2}$
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