Question

Salut, ma puteti ajuta la problema numarul 675, nu ii dau de cap deloc...

Answer 1

[tex]\displaystyle\limit\lim_{n\to\infty}\left(n-\displaystyle \sum_{k=1}^{n} e^{\frac{k}{n^2}\right)=\displaystyle\limit\lim_{n\to\infty} \left (1-e^{\frac{1}{n^2}}+1-e^{\frac{2}{n^2}}+\ldots+1-e^{\frac{n}{n^2}}\right)=\\ \\ =\displaystyle\limit\lim_{n\to\infty} \left( -\dfrac{(e^{\frac{1}{n^2}}-1)}{\frac{1}{n^2}}\cdot \dfrac{1}{n^2}-\ldots - \dfrac{(e^{\frac{n}{n^2}}-1)}{\frac{n}{n^2}}\cdot \dfrac{n}{n^2}\right)=\\ [/tex]

$\displaystyle\limit\lim_{n\to\infty} \left(-\dfrac{1}{n^2}-\dfrac{2}{n^2}-\ldots - \dfrac{n}{n^2}\right)=\left( -\dfrac{n(n+1)}{2n^2}\right)= \boxed{\dfrac{-1}{2}}$