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