@PedroTamaroff heh, my solution is so dumb I think I misunderstood the problem. If the sequence is real and convergent, then surely at least one of three things must be true: there are infinitely many zeros; there are infinitely many positive terms; there are infinitely many negative terms. If there are infinitely many positive terms, take the sequence $x'_1, x'_2,\ldots$ of positive terms. It is absolutely convergent because all the terms are positive and it is convergent.
@PedroTamaroff Suppose a monotone sequence $x_{n_1}, x_{n_2},\ldots$ with $n_1\lt n_2\lt \cdots$ exists. This is convergent, so it is the sequence of partial sums of an absolutely convergent series.
@PedroTamaroff it is a monotone sequence. If it is constant, we are done. If it is increasing, the terms of the series can be written as $S_{m+1}=x_{n_1}+\sum_{k=1}^m x_{n_{k+1}}-x_{n_{k}}$. It is monotone, so $x_{n_{k+1}}-x_{n_k}\geq 0$. This is absolutely convergent. The decreasing case is similar. I think the existence of such monotone sequence can be established by the pigeonhole principle, wait a bit.
@PedroTamaroff I think there is a theorem that says that every sequence of length $2n$ contains a monotone sequence of length $n+1$. This implies the existence. I will try to prove it
@GabrielR. never proved my claim, so I prove it: since $x_n$ is Cauchy, we can produce an increasing sequence $n_1<n_2<n_3<\cdots$ such that $$|x_n-x_{n_k}|<2^{-k}$$ whenever $n>n_k$, in particular then $$|x_{n_{k+1}}-x_{n_k}|<2^{-k}$$
@IanMateus The pre-Q&A question already got me stumped
"Find a polynomial f(x) with all coefficients in {0, 1, ..., 9} such that f(2) is prime but f(x) is reducible (i.e., f(x) = g(x)h(x) with g, h in Z[x])"
"I am a 9-th grader interested in number theory and theory of equations. I am, at present, working on sextic equations and some results related to sum of cubes equal to square of sum property."
Last problem was rather easy. Let's try another one. If $\liminf\limits_{n\to\infty} x_n=0$, there exists infinitely many $n$ such that $x_n<x_k$ for every $k<n$.
Oh, I have something interesting. It's easy with the Axiom of Choice, I guess, is it possible without though? Prove that there exists a set of all sets of Natural numbers with one and only one item in them.
Yeah. Have to prove that the following exists: A set that has within it all the sets of of natural numbers that has within them one and only one number (item).
I may use the Axiom of choice to answer it, so it's easy. Wonder if it's even possible without the Axiom of choice, though.
I need to think about one of my optional homework problems for my diff geo class. One of my students was stuck on it, and I haven't thought about it for years.
Here it is in a nutshell. Suppose $\alpha$ and $\beta$ are curves so that $\beta(s)$ lies on the normal line to $\alpha$ at $s$ and vice versa. Then $\beta = \alpha + rN$ for some constant $r$. The problem is to prove that the product of the torsions of $\alpha$ and $\beta$ at corresponding points is constant.
@KarlKronenfeld yep. We can build such set according to the axiom of partial sets according to a given character/property - character or property is something like < being an order on A, you can prove it with the accepted 'minimal' signs
