# 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.

-
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

## 1 Answer

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
@Justin: The answer to the first question is that I want to show that the sequence converges to 0, and this is equivalent to showing that the sequence of absolute values goes to zero. (In general a sequence {x_n} converge to L if |x_n - L| converges to 0, and here L=0.). –  Jonas Meyer Oct 20 '10 at 1:30
@Justin: An answer to the second question is given by an analogue of the "squeeze theorem" for sequences: en.wikipedia.org/wiki/Squeeze_theorem. The sequence is "squeezed" between -2|a_n| and 2|a_n|, which both go to zero. It's basically because we can make 2|a_n| as small as we want by taking n big enough, and so if we have a sequence {x_n} with |x_n| smaller than 2|a_n|, then we can also make |x_n| as small as we want by taking n big enough. –  Jonas Meyer Oct 20 '10 at 1:35
@Justin: Incidentally, you didn't have to accept the answer if it wasn't clear :) –  Jonas Meyer Oct 20 '10 at 1:36
I tried to undo it but it was not working, something about having accepted it too recently :) Thanks a lot. –  fdart17 Oct 20 '10 at 1:43