Statements possibly equivalent to the $\epsilon{-}N$ definition of limit of a sequence Which of these statements are equivalent to the $\epsilon{-}N$ definition of limit of a sequence?
  
  (a) For every integer $m&gt;0$, there in an integer $N&gt;0$ such that if $n &gt; N$ then 
  $|a_n - \alpha| &lt; \frac1m$
  
  (b) For each $0&lt;\epsilon &lt;1$, there in an integer $N&gt;0$ such that if $n&gt;N$ then $|a_n - \alpha| &lt; 3\epsilon$
  
  (c) For each $0&lt;\epsilon &lt;1$, there in an integer $N&gt;0$ such that if $n&gt;N$ then $|a_n - \alpha| &lt; \frac{1}{\epsilon}$
  
  (d) For each $\epsilon &gt; 0$, there in an integer $N&gt;0$ such that if $n&gt;N$ then $|a_n - \alpha| &lt; \frac{1}{\epsilon}$
  
  (e) For each $\epsilon &gt; 0$, there in an integer $N&gt;0$ such that if $n&gt;N^2$ then $|a_n - \alpha| &lt; \epsilon$
  
  (f) For each $\epsilon &gt; 0$, there in an integer $N&gt;0$ such that if $n=N+1$ then $|a_n - \alpha| &lt; \epsilon$


Can anyone help me with this? To me all of these seem equivalent to the original
$\epsilon-N$ definition where $\alpha$ is the limit of the sequence. In all the cases we find an $N$ such that for all $n&gt;N$ the statement $|a_n - \alpha| &lt; something$ is true. In (e) and (f) $n&gt;N^2$ and $n = N+1$, respectively, but in both cases it is still greater than $n&gt;N$ so I don't see how this effects the statements.
The definition of limit of a sequence $(a_n)_{n\in \mathbb{N}}$ is:
$$\lim_{n \to +\infty} a_n = \alpha$$ means:


  for every $\varepsilon &gt; 0$, "no matter how small", $(a_n)_{n\in \mathbb{N}}$ is, from some point on, within $\varepsilon$ of $\alpha$.


"$x$ is within $\varepsilon$ of $y$" means $|y -x| &lt; e$.

A condition or property is true a sequence $(a_n)_{n\in \mathbb{N}}$ "from some point on" means that there's an $N \in \mathbb{N}$ such that the property or condition is true about $a_n$ for all $n \ge N$. The $a_i$ can fiddle around for some finite number of $i$, but eventually (starting at $N$) the $a_i$ have to settle down and from that point ($N$) on, for all $n \ge N$, $a_n$ fulfills the condition.

Loosely: to within whatever degree of accuracy or precision we require, the sequence eventually gets that close to $\alpha$ and stays that close forever after.

With this in mind, let's run through the statements you list. They're all attempts at expressing that the sequence converges to $\alpha$.

Statements (a) - (f)

(a) True (equivalent). For any real $\varepsilon$ there's an integer $m$ with $\frac 1 m \le \varepsilon$.

(b) True. Sufficiently small $\varepsilon$ will be smaller than $1$. We're interested in shrinking the neighborhood around $\alpha$ and ensuring that no matter how small it is, from some point the on the sequence gets into that neighborhood and stays in that neighborhood.

(c) False (not equivalent): This says that for any arbitrarily large real number $r$ (namely, $1/\varepsilon$), the sequence is eventually within distance $r$ of the limit. But... big deal. This doesn't tell us that the sequence gets as close as we require to $\alpha$ and stays that close forever, which is what convergence means.

(d) True. Unlike (c), $\varepsilon$ can be taken large, so $1/\varepsilon$ can be arbitrarily small.

(e) True: Given $\varepsilon$, there is $M$ (changing variable names) such that from $N = M^2$ on, $a_n$ is within $\varepsilon$ of $\alpha$ for $n &gt; N$.

(f) False: it says "$n = N+1$" which doesn't say anything about the "eventual" behavior of the sequence. (Exercise: find a counterexample.)
The definition of limit of a sequence $(a_n)_{n\in \mathbb{N}}$ is:
$$\lim_{n \to +\infty} a_n = \alpha$$ means:


  for every $\varepsilon &gt; 0$, "no matter how small", $(a_n)_{n\in \mathbb{N}}$ is, from some point on, within $\varepsilon$ of $\alpha$.


"$x$ is within $\varepsilon$ of $y$" means $|y -x| &lt; e$.

A condition or property is true a sequence $(a_n)_{n\in \mathbb{N}}$ "from some point on" means that there's an $N \in \mathbb{N}$ such that the property or condition is true about $a_n$ for all $n \ge N$. The $a_i$ can fiddle around for some finite number of $i$, but eventually (starting at $N$) the $a_i$ have to settle down and from that point ($N$) on, for all $n \ge N$, $a_n$ fulfills the condition.

Loosely: to within whatever degree of accuracy or precision we require, the sequence eventually gets that close to $\alpha$ and stays that close forever after.

With this in mind, let's run through the statements you list. They're all attempts at expressing that the sequence converges to $\alpha$.

Statements (a) - (f)

(a) True (equivalent). For any real $\varepsilon$ there's an integer $m$ with $\frac 1 m \le \varepsilon$.

(b) True. Sufficiently small $\varepsilon$ will be smaller than $1$. We're interested in shrinking the neighborhood around $\alpha$ and ensuring that no matter how small it is, from some point the on the sequence gets into that neighborhood and stays in that neighborhood.

(c) False (not equivalent): This says that for any arbitrarily large real number $r$ (namely, $1/\varepsilon$), the sequence is eventually within distance $r$ of the limit. But... big deal. This doesn't tell us that the sequence gets as close as we require to $\alpha$ and stays that close forever, which is what convergence means.

(d) True. Unlike (c), $\varepsilon$ can be taken large, so $1/\varepsilon$ can be arbitrarily small.

(e) True: Given $\varepsilon$, there is $M$ (changing variable names) such that from $N = M^2$ on, $a_n$ is within $\varepsilon$ of $\alpha$ for $n &gt; N$.

(f) False: it says "$n = N+1$" which doesn't say anything about the "eventual" behavior of the sequence. (Exercise: find a counterexample.)