So I have a problem of
$$a_n = \left(1 + \frac{2}{n}\right)^n$$
I need to determine whether it is diverging or converging and find the limit if it is converging I found an answer on symbol lab of $e^2$ but I do not know how they got that type of answer
It is indeed $e^2$, because of the well-known result that $\lim_{n\to\infty}(1+\dfrac xn)^n=e^x$.
Probably my favorite proof of this uses a single interval Riemann sum, coupled with the sandwich or squeeze theorem for limits.
I will try to recreate it if you like.
To wit, define $\ln x:=\int_1^x\dfrac1t\operatorname{dt}$. Then by doing an upper and lower Riemann sum (one subinterval) we get: $$\dfrac xn\le\ln (1+\dfrac xn)\le\dfrac n{n+x}\cdot\dfrac xn$$. Now take $e$ of everything, raise everything to the $n$th power, and take the limit.
Well, once you know the series converges, that is actually one of the definitions of $e^2.$ Here is another way.
Consider $\ln(a_n)=n\ln\left(1+\frac{2}{n}\right).$ Now we look at the form $$\frac{\ln(1+2/n)}{1/n}=\ln(a_n).$$ We see that $$\frac{(\ln(1+2/n))'}{(1/n)}=\frac{\frac{1}{1+2/n}}{-1/n^2}\cdot\frac{-2}{n^2}=\frac{2}{1+2/n}\to 2.$$ So by L'Hopital's $$\ln(a_n)\to 2.$$ By continuity $$e^{\ln(a_n)}=a_n\to e^2.$$
