4

If $\{x_{n}\}$ is a sequence of positive real numbers, $0<x_{n}<1$ and $\lim_{n\to\infty}x_{n}=0$. We Know that any subsequence of $x_{n}$ will converges to zero, right! Now my question is: Can we find (construct) a subsequence $x'_{n}$ of $x_{n}$ such that $$\lim_{n\to\infty}\frac{x'_{n}}{x_{n}}=x$$ for nonzero $x$.

(For example, if $x_{n}=\frac{1}{n}$, then we can choose $x'_{n}:=x_{2n}=\frac{1}{2n}$ and we get $\lim_{n\to\infty}\frac{x'_{n}}{x_{n}}=1/2$).

Edit: Above I said "for nonzero $x$", and I didn't specified a value for $x$, all I want is just a nonzero limit.

Paul
  • 93
  • True, any subsequence of a convergent sequence will also converge to the same limit. I suspect that the answer to your second question is not in general. There is a rather special uniformity to the example you chose, both in its monotonicity and the way it progresses. I'll think on a counterexample (unless someone beats me to it or comes up with a proof in the meantime). – Cameron Buie Jun 10 '12 at 18:32
  • Ok, as a summary: The result could be true if the sequence $x_{n}$ is increasing or decreasing, right! – Paul Jun 11 '12 at 01:28

3 Answers3

2

I suppose the trivial answer to your question is "yes." After all, one can always take $n' = n$, and then $\lim_{n \rightarrow \infty} x_{n'}/x_n = 1$, since obviously each term is one.

One possible way to make the problem less trivial is to require that $n' > n$ for all $n$. A counterexample to something like this can be given by $x_n = 1/2^{2^n}$. Note that $x_{n+1}/x_n = 2^{2^n - 2^{n+1}} = 1/2^{2^n}$, any such ratio must tend to 0.

Tony
  • 6,718
0

Interesting question, which could get difficult if we put further conditions on $(x_n)$. We could cheat and use a monotonically decreasing sequence $(x_n)$, and $x=3$. But let us instead use an $x \lt 1$.

Let our sequence $(x_n)$ be given by $x_n=\frac{1}{n}$ when $n$ is not a power of $2$, and $x_n=\frac{1}{2^n}$ when $n$ is a power of $2$. So $x_n$ decreases rather slowly most of the time, but occasionally takes a dramatic dip.

Let $x=\frac{1}{3}$, and suppose that we have a subsequence $(x_n')$ such that $$\lim_{n\to\infty}\frac{x_n'}{x_n}=\frac{1}{3}.$$ Let $m=2^k$ be a large power of $2$.

If $m$ is sufficiently large, $\frac{x_{n+1}'}{x_{n+1}}\approx\frac{1}{3}$, so, informally, $x_n'\approx \frac{1}{3(n+1)}$. But then we cannot have $\frac{x_n'}{x_n}$ anywhere near $\frac{1}{3}$.

André Nicolas
  • 507,029
  • Not only that sequence of your takes a dramatic dip once in a while, but it also is NOT monotonically convergent to zero, and this opens up possibilities for that $,x,$ – DonAntonio Jun 10 '12 at 18:51
  • Yes, what makes the example work is the non-monotone nature of the sequence. One can get examples with monotone sequences, as long as the rate of going down is dramatic enough. – André Nicolas Jun 10 '12 at 18:58
0

One thing seems to be sure: if your sequence converges monotonically to zero (from above, since we're given $\,x_n>0\,$) then any subsequence will bound it elementwise from below: $\,\,x'_n\leq x_n\,,\,\forall n\,$, and then any possible limit of the quotient of both will have to be in $\,[0,1]\,$ , so if we have $\,1<x\in\mathbb{R}\,$ we'll have to begin with a seq. that converges to zero non-monotonically, and this already rules out lots of pretty simple and basic examples, and also shows us that either we put some conditions on the sequence $\,\{x_n\}\,$ or else the answer to your question is : no, not any real $\,x\,$ can be gotten as a limit of that quotient for any sequence.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287