A formal proof for an affirmation about sequences

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Homework Statement



Prove that:

The sequence x_n \to x if and only if there is a M > 0 such that \forall \epsilon > 0 , \exists n_\epsilon \in \mathbb{N} and n\geq n_\epsilon we have | x_n - x | < \epsilon M

Homework Equations



The first implication "=>" is proved by choosing M = 1. Then, the problem statement is exactly the epsilon-convergence theorem.

I don't know how to prove "<=".

The Attempt at a Solution



I tried to chose \epsilon = \epsilon_1 = \frac{\epsilon}{M} , and I get that:

\exists n_{\epsilon_1} \in \mathbb{N} and n\geq n_{\epsilon_1} we have | x_n - x | &lt; \epsilon_1 M = \epsilon

So I found an n_{\epsilon_1} such that ... | x_n - x | &lt; \epsilon. But I should've found a rang that depends on epsilon, not on epsilon1. And this confuses me a little because I don't know for sure if I can state the following: I've found n_{\epsilon_1} so, by substituing , I've found n_{\frac{\epsilon}{M}}, which depends on epsilon, but also depends on M, which is a constant, and the theorem is proved. I'm not convinced. Can anyone help me?

P.S: I'm not a native speaker and I definitely have problems with this formal language used, so I hope you guys understood exactly what I said. To me, it seems clear. If it's not, please, let me know, so I can rephrase it. :D
 
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Stating this a little less formally, what you are really trying to show is that given an M the statement

|##x_n## -x| < ##\epsilon##M (1)

for any ##\epsilon## means |##x_n## -x| goes to zero as n gets large.

So you want to show that if (1) is true then for any ##\epsilon## there is some ##N_{\epsilon}##such that for all n > ##N_{\epsilon}##
|##x_n## -x| < ##\epsilon##.

You set up an ##\epsilon_1 = \epsilon##/M which is correct. Now pick ##N_{\epsilon}## so that for all n > N |##x_n## -x| < ##\epsilon_1##M. Your hypothesis says this should be possible. Of course ##\epsilon_1##M = ##\epsilon##, so you have your result.

What you said is either correct or very close to correct. You knew what you were trying to do. The problem is in stating is so that you fulfill your hypothesis and still get your result. I don't think it is easy even if you speak English well.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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