lim-n-e-1-n-e-2-n-e-n-n-n-ho-is-can-you-help-me-step-by-step-

Question Number 110616 by mohammad17 last updated on 29/Aug/20

{l i m}_{n \to \infty} \frac{e^{\frac{1}{n}} + e^{\frac{2}{n}} + \dots \dots + e^{\frac{n}{n}}}{n}

h o i s c a n y o u h e l p m e s t e p b y s t e p

Answered by Dwaipayan Shikari last updated on 29/Aug/20

\lim_{n \to \infty} \frac{1}{n} \underset{r = 1}{\sum^{n}} e^{\frac{r}{n}}

= \int_{0}^{1} e^{x} d x = e - 1

Commented by mohammad17 last updated on 30/Aug/20

s i r {l i m}_{n \to \infty} \frac{1}{n} = 0 \Rightarrow \underset{r = 1}{\sum^{n}} e^{\frac{r}{n}} = 0

Commented by Aziztisffola last updated on 31/Aug/20

\underset{r = 1}{\sum^{n}} e^{\frac{r}{n}} = e^{\frac{1}{n}} + {(e^{\frac{1}{n}})}^{2} + \dots + {(e^{\frac{1}{n}})}^{n}

= e^{\frac{1}{n}} \frac{e - 1}{e^{\frac{1}{n}} - 1} = \frac{e - 1}{1 - e^{\frac{- 1}{n}}}