Answer
converges to ${\dfrac{1}{e}}$.
Work Step by Step
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists.
Consider $a_n=(1+\dfrac{1}{n})^{-n}$
Apply limits to both sides.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(1+\dfrac{1}{n})^{-n}$
Put $y=\lim\limits_{n \to \infty}a_n=(1+\dfrac{1}{n})^{-n}$
Use logarithmic rule: $\ln a^n=n \ln a$
$\ln y=\lim\limits_{n \to \infty}(-n) \ln (1+\dfrac{1}{n})$
$y=\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{0}{0}$
Since, we can see that the limit has the form of $\frac{0}{0}$, so take the help of L-Hospital's rule.
$\ln y=\lim\limits_{n \to \infty}[ -\dfrac{\ln (1+1/n)}{1/n}]$
$\ln y=\lim\limits_{n \to \infty} \dfrac{-1/ (1+1/n)}{1}$
$\ln y=-1 \implies y={\dfrac{1}{e}}$
Therefore, the sequence converges to ${\dfrac{1}{e}}$.
