Prove that $\displaystyle lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]=1$

I am thinking something like that :

$\displaystyle lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]= lim[\frac{n^2}{n^2(n+\frac{1}{n^2})}+\frac{n^2}{n^2(n+ \frac{2}{n^2})}+...+\frac{n^2}{n^2(n+\frac{1}{n})}]$ and since $\displaystyle \frac{1}{n^2}\rightarrow0$ we get

$\displaystyle lim[\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}]=1$

Am i right?