Question: Suppose $(q_{n})_{n=1}^{\infty}$ is a sequence of real numbers such that Lim$_{n \rightarrow \infty} q_{n} = + \infty$. Show that we can find a sequence $(a_{n})_{n=1}^{\infty}$ such that $\sum a_{n}$ is convergent but $\sum a_{n}q_{n}$ is divergent.
My solution is as follows:
Without loss of generality, assume the sequence $(q_{n})_{n=1}^{\infty}$ is a positive sequence. Consider a sequence $(a_{n})_{n=1}^{\infty}$ such that each term in the sequence alternates from a positive term to a negative term (i.e.: the sequence $(a_{n})_{n=1}^{\infty}$ is an alternating series). Let $a_{n}=\frac{1}{q_{n}}(-1)^{n}$. From the question, we know that Lim$_{n \rightarrow \infty} q_{n} = + \infty$. By the alternating series test, the series $\sum a_{n}$ is convergent, since Lim$_{n \rightarrow \infty} \frac{1}{q_{n}} = 0$ and the sequence $\frac{1}{q_{n}}$ is a decreasing sequence since $q_{n}$ is an increasing sequence. So far, we have found a sequence $(a_{n})_{n=1}^{\infty} = \frac{1}{q_{n}} (-1)^{n}$ and proved by the alternating series test that $\sum a_{n}$ is convergent. Now, the series $\sum a_{n}q_{n} = \sum \frac{1}{q_{n}} (-1)^{n}q_{n} = \sum (-1)^{n}$, which is divergent, and so we are done.
However, I was reading over my solution and realised I made a mistake in assuming that the sequence $\frac{1}{q_n}$ is a decreasing sequence because that is not necessarily the case. Also, when I showed someone, they pointed out that an alternating series doesn't necessarily mean it is divergent, it just means it is not convergent since it doesn't tend to positive or negative infinity.
How should I alter my solution to correct this, or if needs be, what is the correct solution?
Thanks
