# Question about the limit comparison test

I'm studying for my analysis midterm and one of the practice study questions provided was the following.

Provide an example of two sequences $x_n$ and $y_n$ such that $\displaystyle\lim_{n\to\infty}\frac{x_n}{y_n} = 1$ but the limit comparison test cannot be applied.

I'm pretty confused about this question. I think the only way the limit comparison test cannot be applied is if not all of the terms are positive. But if you take the limit like this and get 1, then isn't that applying the limit comparison test...? And since you get a constant (1), then doesn't that mean you can apply the test accordingly? I'm really not sure what kinds of sequences could be applied here. Thanks in advance.

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I think the wording is a bit imprecise, and the question is probably trying to get at the fact that the limit comparison test always gives $1$ for a $p$-series, and thus the limit comparison test is always inconclusive for a $p$-series (some of which converge and some of which diverge). "Cannot be applied" seems to me to be a meta-mathematical human description, whereas "inconclusive" is the standard mathematical term for when the test provides no information about convergence. – Dave L. Renfro Mar 10 '14 at 19:32
Regarding my last comment, I was thinking of the ratio and root tests, so my comment doesn't really apply. [I tried editing my comment to $0$ words, thereby effectively canceling my comment, but apparently this is not how one cancels comments (and I see no information on my screen that tells me how to cancel a comment).] Nonetheless, I still think "cannot be applied" is not a good choice of words for the reason I gave earlier. – Dave L. Renfro Mar 10 '14 at 19:37

Let $$x_n=\frac{(-1)^n}{\sqrt n+(-1)^n}\quad n\ge2$$ then $$x_n=\frac{(-1)^n}{\sqrt n}\frac{1}{1+(-1)^n/\sqrt n}=\frac{(-1)^n}{\sqrt n}\left(1-\frac{(-1)^n}{\sqrt n}+O\left(\frac1n\right)\right)$$ so we see that $\displaystyle \sum_n x_n$ is divergent although $x_n\sim_\infty y_n=\frac{(-1)^n}{\sqrt n}$ and the series $\displaystyle \sum_n y_n$ is convergent by Leibniz theorem.