2
$\begingroup$

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.

  • 0
    OK, I'll delete my comments too. If you're sure this is the actual question, I don't need to comment anyway. :)2010-10-19

1 Answers 1

3

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.

  • 0
    I tried to undo it but it was not working, something about having accepted it too recently :) Thanks a lot.2010-10-20