Answer
First five terms are : $\implies n=1$: $\dfrac{(1+1)(1+2)}{2(1)^2}=3$
$\implies n=2$: $\dfrac{(2+1)(2+2)}{2(2)^2}=1.5$
$\implies n=3$: $\dfrac{(3+1)(3+2)}{2(3)^2}=1.111111$
$\implies n=4$: $\dfrac{(4+1)(4+2)}{2(4)^2}=0.9375$
$\implies n=5$: $\dfrac{(5+1)(5+2)}{2(5)^2}=0.84$
Converges to $\dfrac{1}{2}$.
Work Step by Step
Plugging $n = {1,2,3,4,5}$ in $\dfrac{(n+1)(n+2)}{2n^2}$.
$\implies n=1$: $\dfrac{(1+1)(1+2)}{2(1)^2}=3$
$\implies n=2$: $\dfrac{(2+1)(2+2)}{2(2)^2}=1.5$
$\implies n=3$: $\dfrac{(3+1)(3+2)}{2(3)^2}=1.111111$
$\implies n=4$: $\dfrac{(4+1)(4+2)}{2(4)^2}=0.9375$
$\implies n=5$: $\dfrac{(5+1)(5+2)}{2(5)^2}=0.84$
We see that $\lim\limits_{n \to \infty} \dfrac{(n+1)(n+2)}{2n^2}=\lim\limits_{n \to \infty} \dfrac{n^2+3n+2}{2n^2}\\=\lim\limits_{n \to \infty} \dfrac{1}{2}+\lim\limits_{n \to \infty} \dfrac{3}{2n}+\lim\limits_{n \to \infty} \dfrac{1}{n^2}\\=\dfrac{1}{2}+0+0\\=\dfrac{1}{2}$
Therefore, the given series converges to $\dfrac{1}{2}$.
