Answer
Converges to $1$.
Work Step by Step
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists.
Consider $a_n= \sqrt[n] {2n+1}$
or, $a_n= (2n+1)^{1/n}$
Apply limits to both sides.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty} (2n+1)^{1/n}$
Let us consider $y=\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty} (2n+1)^{1/n}$
Use logarithmic rule: $\ln a^n=n \ln a$
$\ln y=\lim\limits_{n \to \infty} \ln (2n+1)^{1/n}$
$\ln y=\lim\limits_{n \to \infty} \dfrac{\ln (2n+1)}{n}$
Since, we can see that the limit has the form of $\frac{\infty}{\infty}$, so take the help of L-Hospital's rule.
$\ln y=\lim\limits_{n \to \infty} (\dfrac{2}{2n+1})(\dfrac{1}{1})$
$\ln y=0$
$e^{\ln y}=e^0 \implies y=1$
Therefore, the sequence converges to $1$.
