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Show that $(x_n)=(\frac{n}{n^2+1})$ converges to zero.

Here's the definition of convergence:

$$\forall \epsilon > 0, \exists N \in N s.t (n\geq N) \implies |{a_n-L}|<\epsilon$$

By solving I get $$\frac{1}{n}<\epsilon$$

Why should I choose the N to be the ceiling of $\frac{1}{\epsilon}$ and not floor

Wang YeFei
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Mengen Liu
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    A standard approach would be $$ \frac{n}{n^2+1}\leqslant \frac{n}{n^2}=1/n < \epsilon$$ if and only if $n> \frac{1}{\epsilon}$ so given $\epsilon > 0$ choose $N>\frac{1}{\epsilon}$ – jcneek Oct 11 '22 at 22:23

2 Answers2

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You are trying to show that you can choose a value of $N$ that makes the convergence statement true for every $n > N$. As you've shown, this reduces to finding a value of $N$ for which $n \geq N \implies \frac{1}{n} < \epsilon$. We can re-arrange that inequality to become $n > \frac{1}{\epsilon}$, and since it's true for all $n \geq N$ it has to be true for $n = N$, so we need $N > \frac{1}{\epsilon}$. But $\lfloor \frac{1}{\epsilon} \rfloor \leq \frac{1}{\epsilon}$, so taking the floor doesn't satisfy our requirements. On the other hand, $\lceil \frac{1}{\epsilon} \rceil \geq \frac{1}{\epsilon}$, so we're good to go from there on.

(Technically there's some fiddling around with strict and non-strict inequalities you have to check too, but that's not a big problem.)

ConMan
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It should be a comment, but I made it an answer. Because the ceiling of $\dfrac{1}{\epsilon} > \dfrac{1}{\epsilon}$ while the floor is not.

Wang YeFei
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