0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

You started with a WLOG reduction - assume that the $q_n$ are positive. How do we do that reduction if it isn't? We just assign $a_n=0$ whenever $q_n$ is negative. Then those terms contribute zero to the series of products. If we drop them from the sum and reindex, we've effectively passed to the subsequence of positive terms in the $a_n$.

That reduction cut out finitely many terms. We can do the same on a larger scale, and pass to any subsequence, no matter how sparse (as long as it's still infinite). So then, we find a subsequence that increases to $\infty$, and pass to that by choosing $a_n$ to be zero everywhere else. Drop the zeros and reindex, then apply your solution to that reduced sequence.

This is, of course, not the only way.

$\endgroup$
0
$\begingroup$

Define $$n_i=\text{The least positive integer }n\text{ for which }q_{n}>3^i$$and$$a_n=\begin{cases}{1\over 2^n}&,\quad n=n_i \text{ for some }i\\0&,\quad \text{otherwise}\end{cases}$$then $\sum_{n=1}^\infty a_n$ is obviously convergent, while $a_nq_n$ grows with a speed of at least $\left( {3\over 2}\right)^n$ at $n_i$s forever and makes the corresponding series divergent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy