Answer
First five terms are: $-1; 0; (\dfrac{1}{3})^3; (\dfrac{2}{4})^4; (\dfrac{3}{5})^5$
Converges to $e^{-2}$.
Work Step by Step
Plugging $n = {1,2,3,4,5}$ in $(1-\dfrac{2}{n})^n$.
$\implies n=1$: $(1-\dfrac{2}{1})^1=-1$
$\implies n=2$: $(1-\dfrac{2}{2})^2=0$
$\implies n=3$: $(1-\dfrac{2}{3})^3=(\dfrac{1}{3})^3$
$\implies n=4$: $(1-\dfrac{2}{4})^4=(\dfrac{2}{4})^4$
$\implies n=5$: $(1-\dfrac{2}{5})^5=(\dfrac{3}{5})^5$
We see that $l=\lim\limits_{n \to \infty} (1-\dfrac{2}{n})^n$
or, $\ln l=\lim\limits_{n \to \infty} x \ln (1-\dfrac{2}{n})\\=\lim\limits_{n \to \infty} \dfrac{-2n}{n-2}\\=\lim\limits_{n \to \infty} \dfrac{-2}{1-2/n}$
or, $\ln l=-2$
or, $ l=e^{-2}$
Therefore, the given series converges to $e^{-2}$.
