Answer
Converges to $1$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt [n] {n^2+n}$
Since, $ e^{\ln x}=x$
So, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt [n] {n^2+n}=\lim\limits_{n \to \infty} e^{\frac{\ln [n(n+1)]}{n}}$
or, $=e^{\lim\limits_{n \to \infty} \frac{2n+1}{n^2+n} (\frac{1/n^2}{1/n^2})}$
or, $=e^0$
Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is Convergent and converges to $1$
