Answer
Sequence converges to $\pi$
Work Step by Step
$\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\sqrt n sin(\frac{\pi}{\sqrt n})$
$=\lim\limits_{n \to \infty}\frac{sin(\frac{\pi}{\sqrt n})}{\frac{1}{\sqrt n}}$
Substitute $\frac{1}{\sqrt n}=t$
When $nā\infty$, $\frac{1}{\sqrt n}ā0$
Therefore,
$\lim\limits_{n \to \infty}\frac{sin(\frac{\pi}{\sqrt n})}{\frac{1}{\sqrt n}}=\lim\limits_{t \to 0}\frac{sin(\pi t)}{t}$
The limit is in the form $\frac{0}{0}$
Use L'Hospital's Rule
$\lim\limits_{n \to \infty} a_{n}=\lim\limits_{t \to 0}\frac{\pi cos(\pi t)}{1}=\frac{\pi cos(0)}{1}=\pi$
Sequence converges to $\pi$
