Answer
$a_1=\left(1+\frac{1}{1}\right)^1=2$
$a_2=\left(1+\frac{1}{2}\right)^2=2.25$
$a_3=\left(1+\frac{1}{3}\right)^3=2.37$
$a_4=\left(1+\frac{1}{4}\right)^4=2.441$
$a_5=\left(1+\frac{1}{5}\right)^5=2.488$
$a_6=\left(1+\frac{1}{6}\right)^6=2.522$
The sequence converges and it converges to e(=2.718..).
Work Step by Step
$\lim\limits_{n \to \infty} \left(1+\frac{1}{n}\right)^n =e^{ \lim\limits_{n \to \infty}n \ln(1+\frac{1}{n})}$
Let's first calculate
$ \lim\limits_{n \to \infty}n \ln(1+\frac{1}{n})= \lim\limits_{n \to \infty} \frac{\ln(1+\frac{1}{n}) }{1/n} = \lim\limits_{n \to 0}\frac{\ln(1+n)}{n}$
By L'Hospital's rule, this limit is equal to
$\lim\limits_{n \to 0} \frac{{1}/{(1+n)}}{1} = 1$
Therefore, the limit $\lim\limits_{n \to \infty} S_n = e^{ \lim\limits_{n \to \infty}n \ln(1+\frac{1}{n})} = e^1 =e $
