Limit 1

[LaçinCanAtış]

$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\:e^{\frac{k}{n}}}{n}\right)$

[ArtOfMathSolving]

Cevabımız $\boxed{e-1}$

$\begin{align*} \lim _{n\to \infty }\left(\dfrac{\sum _{k=1}^n\:e^{\frac{k}{n}}}{n}\right)= \lim_{n\rightarrow \infty}\left( \dfrac{(1-e^{\frac{n+1}{n}})}{n(e^{1/n}-1)}\right)=\lim_{n\rightarrow \infty} \left(\dfrac{(e-1)(e^{1/k})}{ne^{1/n-n}}\right)=(e-1)\lim_{n\rightarrow \infty}\dfrac{1}{ne^{1/n}-n}  \end{align*}$

$\text{L'Hopital}$ uygularsak, $\begin{align*}(e-1)\lim_{n\rightarrow \infty}\dfrac{1}{ne^{1/n}-n}=(e-1)\lim_{n\rightarrow \infty}e^{-1/n}=e-1\end{align*}$ buluruz. $\blacksquare$

[ahmetbazın]

Riemann integrali

integralin sınırları 0 dan 1  e üzeri x dx