Answer
$1;1.1622777;1.2426407; 1.2915026;1.3245553$
Converges to $\frac{3}{2}$.
Work Step by Step
Plugging $n = {1,2,3,4,5}$ in $\sqrt {n^2+3n}-n$.
$\implies n=1$: $\sqrt {(1)^2+3(1)}-1=1$
$\implies n=2$: $\sqrt {(2)^2+3(2)}-2=1.1622777$
$\implies n=3$: $\sqrt {(3)^2+3(3)}-3=1.2426407$
$\implies n=4$: $\sqrt {(4)^2+3(4)}-4=1.2915026$
$\implies n=5$: $\sqrt {(5)^2+3(5)}-5=1.3245553$
We see that
$\lim\limits_{n \to \infty} (\sqrt {n^2+3n}-n)=\lim\limits_{n \to \infty} \dfrac{3n}{\sqrt {n^2+3n}+n}\\=\lim\limits_{n \to \infty} \dfrac{3}{\sqrt{1+3/n}+1}=\dfrac{3}{2}$
Therefore, the given series converges to $\frac{3}{2}$.
