Let sequences $\left( a_{n}^{(k)} \right)^{\infty}_{n=1}$, where $k \in \mathbb{N}$, so there are is a finite number of sequences, be unbounded. Must there exist a sequence $\left( X_{n} \right)^{\infty}_{1}$ such that $X_{n}>0$ and $\lim{X_{n}} = 0$ and all

1) $\left( X_{n}a_{n}^{(k)}\right)^{\infty}_{n=1}$ converge?

2) $\left( X_{n}a_{n}^{(k)}\right)^{\infty}_{n=1}$ diverge?

The easy intuitive example is $a_n^{(k)}=n^{k}$ and $X_n = \frac{1}{n^{k}}$ Then $X_n > 0$, $X_n$ converges to 0, and $\left( X_{n}a_{n}^{(k)}\right)^{\infty}_{n=1}$ converges (to 0 or to 1). So, what I have tried so far is this, let $ X_n = \frac{1}{\max(1, \lvert a_n^{(1)} \rvert, \lvert a_n^{(2)} \rvert,\dots, \lvert a_n^{(k)} \rvert)^2}$. Then we have $X_n > 0$ and $\lim{X_n}=0$. But I am not sure where to go from now, intuitively, the sequence $\left(X_n a_n^{(k)}\right)^{\infty}_{n=1}$ converges to 0, because $X_n$ is always "decreasing faster", but I am not sure how to prove that this is true and that a sequence like this always exists.

The second part seems easier at first, but I am at even a greater loss there. I cannot think of an example where it would hold.

Anyone could give me some advice on how to proceed?

  • 2
    $\begingroup$ One important clarification is needed here, because the way you phraze your question is slightly ambiguous: Are there an infinite number of sequences indexed by $k$, or are there just $k$ sequences total? (I would interpret the first sentence meaning there are an infinite number of them, but you seem to have assumed in your reasoning that there are only $k$ of them.) $\endgroup$ – Arthur Oct 25 '13 at 10:32
  • $\begingroup$ The way I understood things (as my professor stated them) it is meant as k sequences in total, where $k \in \mathbb{N}$ of course. So I take it there is a finite number of unbounded sequences. $\endgroup$ – Pr0bability Oct 25 '13 at 13:02
  • $\begingroup$ Then I would suggest something like $$ X_n = \prod_{i= 1}^k \frac{1}{a_n^{(i)}} $$ for the first one, and perhaps $$ X_n = \frac{1}{\log \min_i\left(a_n^{(i)}\right)} $$ for the second. None of these work in general for an infinite number of sequences, though. As a side note, if $k$ is fixed, then writing $a_n^{(k)}$ specifically means the last sequence. If you just want one non-specified sequence, you should use another index to signify which one (like I did with $i$ above). $\endgroup$ – Arthur Oct 25 '13 at 13:09
  • $\begingroup$ Hmm, we cannot use log yet (because we have not yet defined an exp function). But thanks for the pointers. $\endgroup$ – Pr0bability Oct 25 '13 at 14:53

If there is only one sequence $(a_n)$, then $x_n=1/(n(1+|a_n|))$ is such that $x_n\gt0$, $x_n\lt1/n$ and $|x_na_n|\lt1/n$ hence $x_n\to0$ and $x_na_n\to0$.

Likewise, let $A_n=\max\{|a_m|;m\leqslant n\}$, then by hypothesis $A_n\to\infty$ hence $x_n=1/(1+\sqrt{A_n})$ is such that $x_n\gt0$, $x_n\to0$, and, each time $(|a_n|)$ reaches a new maximum, $|a_n|=A_n$ hence, if $A_n\gt1$ then $|x_na_n|=A_n/(1+\sqrt{A_n})\gt\sqrt{A_n}/2$, in particular, $(x_na_n)$ is unbounded.

If there are $k$ sequences $(a^{(i)}_n)$ for $1\leqslant i\leqslant k$, assume that, for each $i$, $(x^{(i)}_n)$ is such that $x_n^{(i)}\gt0$, $x_n^{(i)}\to0$ and $x_n^{(i)}a_n^{(i)}\to0$. Let $x_n=\min\{x_n^{(i)};1\leqslant i\leqslant k\}$, then $x_n\gt0$, $x_n\to0$ and $x_na_n^{(i)}\to0$ for each $1\leqslant i\leqslant k$.

Likewise, assume that, for each $i$, $(x^{(i)}_n)$ is such that $x_n^{(i)}\gt0$, $x_n^{(i)}\to0$ and $(x_n^{(i)}a_n^{(i)})$ is unbounded. Let $x_n=\max\{x_n^{(i)};1\leqslant i\leqslant k\}$, then $x_n\gt0$, $x_n\to0$ and $(x_na_n^{(i)})$ is unbounded, for each $1\leqslant i\leqslant k$.

  • $\begingroup$ ...each time $(|a_n|)$ reaches a new maximum, $|a_n|=A_n$ hence, if $A_n\gt1$ then $|x_na_n|=A_n/(1+\sqrt{A_n})\gt\sqrt{A_n}/2$, in particular, $(x_na_n)$ is unbounded. That is not right. In general, $|a_n| \leq A_n$. $\endgroup$ – Pr0bability Nov 4 '13 at 18:00
  • $\begingroup$ Yes. And? What exactly "is not right"? $\endgroup$ – Did Nov 4 '13 at 18:04
  • $\begingroup$ Well, you go from $|x_na_n|=|a_n/(1+\sqrt{A_n})|$ to $|x_na_n|=A_n/(1+\sqrt{A_n})$. Which I think is wrong. I think it should go like this: $|x_na_n|=|a_n/(1+\sqrt{A_n})|\geq|a_n|/(1+\sqrt{|a_n|}) > |a_n|/2$. I hope that is right. $\endgroup$ – Pr0bability Nov 4 '13 at 18:20
  • $\begingroup$ Except that one writes this only for the times $n$ such that $(|a_n|)$ reaches a new maximum. At these specific times, $|a_n|=A_n$. $\endgroup$ – Did Nov 4 '13 at 18:42

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.