I´m struggling to understand the following notation as an example of a convergent sequence:

The sequence $(a_n)_{n \in\mathbb{N}}$ with $a_n$:= $\frac{1}{n+1}$ converges to $0$. This is because: For $\epsilon > 0$ define $n_0$:= $\lfloor \epsilon^{-1} \rfloor$. Then the following holds for all $n\geq n_0$:

$\left\vert a_n-0\right\vert = \left\vert \frac{1}{n+1}-0\right\vert =\frac{1}{n+1}\leq \frac{1}{n_0+1}=\frac{1}{\lfloor \epsilon^{-1}\rfloor+1}\leq\frac{1}{\epsilon^{-1}}= \epsilon$

I know the formal definition of limits and the role of $\epsilon$. However I dont understand this notation $\epsilon^{-1}$, perhaps because of the negative exponent,as well as the role of that $\lfloor \epsilon^{-1}\rfloor$, perhaps because of the gaussian brackets. Could somebody be so kind an explain it to me?


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    $\begingroup$ $\epsilon^{-1} = \frac{1}{\epsilon}$. The point is that $$\lfloor \epsilon^{-1}\rfloor + 1 > \frac{1}{\epsilon}.$$ The floor is taken to obtain an integer $n_0$. $\endgroup$ – Daniel Fischer Jan 13 '14 at 16:06
  • $\begingroup$ still dark but it gets brighter :) $\endgroup$ – Googme Jan 13 '14 at 16:09
  • $\begingroup$ @DanielFischer ... so the last $\le$ can in fact be replaced with $<$. (Which is good because formally at this level we want $|a_n-0|<\epsilon$, not just $|a_n-0|\le\epsilon$) $\endgroup$ – Hagen von Eitzen Jan 13 '14 at 16:11
  • $\begingroup$ Daniel, could you please bit lower the abstraction level regarding the gaussian brackets...i still dont fully understand $\endgroup$ – Googme Jan 13 '14 at 16:14
  • $\begingroup$ The gaussian brackets means "the greatest integer lower than". So it's the unique integer $n_0$ such as $n_0\leq 1/\epsilon < n_0+1$. $\endgroup$ – D.L. Jan 13 '14 at 16:28

To show that $a_n \to 0$, for every $\epsilon > 0$, one needs to find an $n_0$ such that $\lvert a_n - 0\rvert = \lvert a_n \rvert < \epsilon$ for all $n \geqslant n_0$. Since all the $a_n$ are positive here, we can forget about the modulus, and need to achieve

$$a_n = \frac{1}{n+1} < \epsilon.$$

That is fulfilled if and only if

$$\frac{1}{\epsilon} < n+1;\qquad\text{or equivalently}\qquad \frac{1}{\epsilon}-1 < n.\tag{1}$$

Now it is clear that if that inequality holds for some $n_0$, then it also holds for all larger $n$, so one needs to find the smallest (positive) integer that satisfies $(1)$, that is

$$n_0 \leqslant \frac{1}{\epsilon} < n_0 + 1.\tag{2}$$

$(2)$ says that $n_0$ is the largest integer that is not greater than $\frac{1}{\epsilon}$, and that means

$$n_0 = \left\lfloor \frac{1}{\epsilon}\right\rfloor.$$

To save vertical space, instead of $\frac{1}{\epsilon}$, we can also write $\epsilon^{-1}$, that is of no mathematical significance.

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