Answer
$$1$$
Work Step by Step
Let us consider an infinite series $\Sigma_{n=k}^{n=\infty} a_n$. This series will converge to the sum $S$ when the sequence of its partial sums $\{S_n\}$ converge to $S$.
That is, $S=\lim\limits_{n \to \infty} S_n$ and $S=\Sigma_{n=k}^{n=\infty} a_n$.
This series will diverge when the limit does not exist and the series will diverge to infinity when the limit is infinite.
Here, we have the sum $S$ is the limit of the sequence of partial sums which is equal to:
$S=\lim\limits_{n \to \infty} \sin \dfrac{(n+1) \pi}{2n+1}\\=\sin \lim\limits_{n \to \infty} \dfrac{(n+1) \pi}{2n+1}\\=\sin \lim\limits_{n \to \infty} \dfrac{(1+1/n) \pi}{2+1/n}$
Because for all $n \gt 0$ , $\lim\limits_{x \to \infty} x^{-n}=0$ and $\lim\limits_{x \to -\infty} x^{n}=\infty$
Thus, $S=\sin \dfrac{(1+0) \pi}{2+0}=\sin \dfrac{\pi}{2}=1$