Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 1:28

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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 $\frac{1}{3}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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