Answer
converges to $1$
Work Step by Step
As we know that $ e^{\ln x}=x$
Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt [n] {n^2+n}$
This implies that $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} e^{\dfrac{\ln [n(n+1)]}{n}}$
or, $e^{\lim\limits_{n \to \infty} (\dfrac{2n+1}{n^2+n}) (\dfrac{1/n^2}{1/n^2})}=0$
Thus, {$a_n$} is Convergent and converges to $1$.
