True or False?
If $\displaystyle \sum_{n=1}^{\infty}a_{n}^2$ is convergent, and$\displaystyle \sum_{n=1}^{\infty} (\frac{\pi}{2} - b_{n})$ is divergent, then $\displaystyle {a_{n}(\sin(b_{n})-1)}$ is a convergent sequence.
So we know that $(a_{n})^2$ is convergent. Therefore, $\lim_{n \to \infty} a_{n}^2 = 0$ I'm not sure what this tells us about or how to relate this to $(a_{n})$.
Also, we can see that since $-1 \ge \sin(b_{n}) \le 1$ And therefore, $-2 \ge \sin(b_{n}) - 1 \le 0$
However, again since I don't know what I can conclude about $(a_{n})$, I'm not sure how I can use this fact.