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Ok. I am trying to solve an exercise in my last calculus assignment, which is the following:

Find an example of two convergent sequences $x_n$ and $y_n$, where $x_n < y_n$, for all $n \in \mathbb{N}$, but $\lim x_n \not < \lim y_n$

Ok, first I though that $x_n$ could be $n^2$ and $y_n$ could be $n^3$, but I immediately noticed this is wrong, because $n^2$ and $n^3$ diverge.

Then, I tried to think about sequences that converge to $0$, since $0$ is simple. So, I came out with $\frac{1}{n}$ and $\frac{1}{n^2}$, but this is also wrong, because for $n = 1$ (the first element), we have that $x_1 = y_1$, and this goes agains the requirements.

Then, I simply though that if I add $1$ to the $n$ at the denominator, I solve easily the problem, so the sequences would be $x_n = \frac{1}{n + 1}$ and $y_n = \frac{1}{(n+1)^2}$.

Ok, but this is stupid because $\frac{1}{(n+1)^2} < \frac{1}{n + 1}$. So at least I should swap $x_n$ and $y_n$...

Am I right?

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Yes, you are right. There are plenty of examples:

  • $y_n=\frac1n$, $x_n=\frac1{2n}$
  • $y_n=\frac1n$, $x_n=\frac1{n+1}$
  • $y_n=2^{-n}$, $x_n=3^{-n}$

...

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Look at $x_n = 1/n$ and $y_n = 2/n$. Then $x_n < y_n$ for all $n$ and $\lim x_n = \lim y_n = 0$.

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You're almost there - switch the $x$s and $y$s. :P

But there's a simpler approach: say, take $x_n=0$ and $y_n={1\over n}$. Never underestimate the value of the trivial example - in this case, a trivially convergent sequence.

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  • $\begingroup$ In fact, as silly as this might sound, one of the key skills to develop in any field of math is how to usefully deploy the trivial examples. $\endgroup$ – Noah Schweber Mar 17 '15 at 19:51
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And there's always sequences with $x_n\nearrow L$ and $y_n \searrow L$. For example, $x_n = -\frac1n$ and $y_n = \frac1n$.

Of course, the conditions on $x_n$ and $y_n$ will always guarantee that $\lim x_n \le \lim y_n$, so the only thing that can happen if $\lim x_n \not < \lim y_n$ is that $\lim x_n = \lim y_n$.

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