Two statements about convergence and showing one implies the other (epsilon-N) Let ${a_n}$ be a sequence of real numbers and let $X$ be a real number. Show that (1) implies (2).
(1) $\exists C >0,\forall k \in \mathbb{N},\exists n_0 \in \mathbb{N},\forall n \in \mathbb{N},n\geq n_0 \implies \lvert a_n- X\rvert \leq \cfrac{C}{k}$
(2) $\forall k \in \mathbb{N}, \exists C >0, \exists n_0\in \mathbb{N}, \forall n\in \mathbb{N}, n\geq n_0 \implies \lvert a_n- X\rvert \leq \cfrac{C}{k}$
My attempt
Let (1) be true. This implies $a_n$ converges to $X$.
Suppose we fix $C=1$. Then for all $k\in \mathbb{N}$, $\cfrac{C}{k}=\cfrac{1}{k}>0$.
So let $\epsilon = \cfrac{1}{k}$. Since lim $a_n=X$, there exists $n_0$ such that for all $n\in \mathbb{N}$, $n\geq n_0$ implies
$$\lvert a_n-X\rvert < \epsilon=\cfrac{1}{k}=\cfrac{C}{k}$$
Which implies
$$\lvert a_n-X\rvert\leq \cfrac{C}{k}$$
Hence for all $k \in \mathbb{N}$, there exists $C>0$, namely $C=1$, such that the rest of (2) follows.

How is this proof? The nested quantifiers were making me lost on what exactly I needed to show so I'm not sure this is correct at all.
I also don't understand the differences in meaning between (1) and (2) regarding convergence. Could anyone explain in simple terms how (2) differs from (1)?
Thank you in advance.
 A: The proof looks fine to me. Since you showed (1) implies convergence to $X$, you can actually set $C$ to any positive real, and the rest will still work. So you showed (1) implies convergence to $X$ implies (2). In particular if the sequence does not converge to $X$, then (1) fails.
But a non-converging sequence can satisfy (2). One example is the following.
Let $a_n\,=\,(-1)^n$. Let $X$ be any real number. Let $\delta$ be the larger of $|X+1|$ and $|X-1|$.
Then the pair $a_n$,$\,X$ satisfies (2). For any given $k\geq1$, assign $C(k)=k \delta.\,\,$
Then $$|a_n\,-\,X|\,\,\leq\,\,\frac{C(k)}{k}$$
holds not just ultimately, but for every $n\geq1$.
If you want to pursue this further, the next question to ask is whether some $a_n$ can be unbounded while satisfying (2).
A: In the first you are given a $C = C_0$ (may not be the only option) that satisfies the conclusion for every $k$, and notice how I named it $C_0$, where it does not depend on $k$.
In the second statement, someone chooses $k$ for you, and you have to pick a $C$ that will work for that $k$; in other words, $C$ may depend on $k$ in the second statement. However, since you have statement 1, the $C$ that you choose could just be the $C$ that works in statement 1.
In your proof, I would not say fix $C = 1$. In fact, you want to fix $C$ to the $C_0$ from the first statement. But this is all encapsulated in the line you wrote well which was "Let $\epsilon = \frac{1}{k}$" which should instead be "Let $\epsilon = \frac{C_0}{k}$..."
Another thing is that you would want to show how you chose $n_0$ in the second part of the proof. In particular, you chose $n_0$ the same way as the first statement.
All of this is just formalization, but you have the right idea of proof. If you need more clarification, I can write up a sample proof, if you want!
EDIT: I should mention: When a statement begins with $\forall$, you generally want to start your proof with "Given..." whatever the $\forall$ is applied to.
