Given a convergent $\{(a_{n})^2\}$, divergent $\{b_{n}\}$, prove $\{a_{n}(\sin(b_{n})-1)\}$ converges

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.

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I am sure there is a better title available... :P – Mariano Suárez-Alvarez Oct 19 '10 at 21:40
OK, I'll delete my comments too. If you're sure this is the actual question, I don't need to comment anyway. :) – Jonas Meyer Oct 19 '10 at 22:10

In the current form of the question, the condition on $b_n$ is irrelevant.
As you said, $a_n^2\to0$. This implies that $a_n\to0$. Then because $|a_n(\sin(b_n)-1)|=|a_n||\sin(b_n)-1|$ and, for the reason you indicated, $|\sin(b_n)-1|\leq2$ for all $n$, the $n^{th}$ term in the sequence has absolute value less than or equal to $2|a_n|$, which converges to 0.
Why are you taking the absolute value of $|a_n(\sin(b_n)-1)|$? Also, how do we know that something smaller than $2|a_n|$ converges to 0? – fdart17 Oct 20 '10 at 1:07