Answer
First five terms are: $\dfrac{4}{2}; (\dfrac{5}{3})^2; (\dfrac{6}{4})^3; (\dfrac{7}{5})^4; (\dfrac{8}{6})^5$
Converges to $e^2$.
Work Step by Step
Plugging $n = {1,2,3,4,5}$ in $(\dfrac{n+3}{n+1})^n$.
$\implies n=1$: $(\dfrac{1+3}{1+1})^1=\dfrac{4}{2}$
$\implies n=2$: $(\dfrac{2+3}{2+1})^2=(\dfrac{5}{3})^2$
$\implies n=3$: $(\dfrac{3+3}{3+1})^3=(\dfrac{6}{4})^3$
$\implies n=4$: $(\dfrac{4+3}{4+1})^4=(\dfrac{7}{5})^4$
$\implies n=5$: $(\dfrac{5+3}{5+1})^5=(\dfrac{8}{6})^5$
We see that $l=\lim\limits_{n \to \infty} (\dfrac{n+3}{n+1})^n$
or, $\ln l=\lim\limits_{n \to \infty} x \ln (\dfrac{n+3}{n+1})\\=\lim\limits_{n \to \infty} \dfrac{2n^2}{(n+1)(n+3)})$
or, $\ln l=2$
or, $ l=e^2$
Therefore, the given series converges to $e^2$.
