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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?

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    $\begingroup$ 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.) $\endgroup$ Commented Oct 25, 2013 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. $\endgroup$ Commented Oct 25, 2013 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). $\endgroup$ Commented Oct 25, 2013 at 13:09
  • $\begingroup$ Hmm, we cannot use log yet (because we have not yet defined an exp function). But thanks for the pointers. $\endgroup$ Commented Oct 25, 2013 at 14:53

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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 $1\leqslant i\leqslant k$.

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  • $\begingroup$ ...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, $|a_n| \leq A_n$. $\endgroup$ Commented Nov 4, 2013 at 18:00
  • $\begingroup$ Yes. And? What exactly "is not right"? $\endgroup$ Commented Nov 4, 2013 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. $\endgroup$ Commented Nov 4, 2013 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, $|a_n|=A_n$. $\endgroup$ Commented Nov 4, 2013 at 18:42

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