Question

$\underset{n\to \infty }{lim}\left(\right(\frac{n}{n}{)}^n+(\frac{n-1}{n}{)}^n+..+({\frac{1}{n}}^n\left)\right)$equals

• A. $\frac{1}{e-1}$
• B. $\frac{1}{1+e}$
• C. $\frac{1}{1-e}$
• D. $\frac{e}{e-1}$

Answer 1

The correct option is D $\frac{e}{e-1}$

$\underset{n\to \infty }{lim}\left(\right(1-\frac{0}{n}{)}^n+(1-\frac{1}{n}{)}^n+..+(1-{\frac{2}{n}}^n)+...)$

$=e^{\underset{n\to \infty }{lim}\left[n\right(\frac{-0}{n}\left)\right]}$ + $e^{\underset{n\to \infty }{lim}\left[n\right(\frac{-1}{n}\left)\right]}$ + $e^{\underset{n\to \infty }{lim}\left[n\right(\frac{-2}{n}\left)\right]}$+...= e${}^0$ + e${}^{-1}$ + e${}^{-2}$ + ...

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