от man111 » 30 Ное 2019, 17:01
Thanks friends i have tried like this way
[tex]\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}=\lim_{n\rightarrow \infty}n^2\sum^{n}_{k=1}\frac{1}{n^2}\cdot \bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}[/tex]
Using limit as a sum
[tex]\displaystyle \lim_{n\rightarrow \infty}n^2\int^{\frac{1}{n}}_{0}x^{x+1}dx[/tex]
Using D , L Hopital Rule
[tex]\displaystyle \lim_{n\rightarrow \infty}\frac{1}{-2n^{-3}}\cdot \bigg(\frac{1}{n}\bigg)^{\frac{1}{n}+1}\cdot \frac{1}{-n^2}=\frac{1}{2}.[/tex]