Answer
Converges to 0.
Work Step by Step
$\lim\limits_{n \to \infty}a_{n}$
$\lim\limits_{n \to \infty}\frac{n(2n-1)}{(2n)^{n}}=\lim\limits_{n \to \infty}\frac{n \times n(2-\frac{1}{n})}{2^{n} \times n^{n}}$
$=\lim\limits_{n \to \infty} \frac{n^{2}(2-\frac{1}{n})}{2^{n} \times n^{n}}$
$=\lim\limits_{n \to \infty} \frac{2-\frac{1}{n}}{2^{n} \times n^{n}}$
$=\lim\limits_{n \to \infty}(2-\frac{1}{n}) \times \lim\limits_{n \to \infty} \frac{1}{2^{n}} \times \lim\limits_{n \to \infty} \frac{1}{n^{n-2}}$
$=(2-0) \times 0 \times 0$
$=0$
Thus, the given sequence converges to 0.
