Answer
Converges
Work Step by Step
Here, we are given that $a_1=1; a_{n+1}=\dfrac{\sin n+1}{\sqrt n} a_n$
Thus, $\dfrac{a_{n+1}}{a_n}=\dfrac{\sin n+1}{\sqrt n} $
Next, we have: $\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} \dfrac{\sin n+1}{\sqrt n} =0 \lt 1$
Hence, we can conclude that the given series converges.
