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DorelXD
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Homework Statement
Prove that:
The sequence [itex] x_n \to x [/itex] if and only if there is a [itex] M > 0 [/itex] such that [itex]\forall \epsilon > 0[/itex] , [itex]\exists n_\epsilon \in \mathbb{N} [/itex] and [itex]n\geq n_\epsilon [/itex] we have [itex] | x_n - x | < \epsilon M [/itex]
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 [itex] \epsilon = \epsilon_1 = \frac{\epsilon}{M}[/itex] , and I get that:
[itex]\exists n_{\epsilon_1} \in \mathbb{N} [/itex] and [itex]n\geq n_{\epsilon_1} [/itex] we have [itex] | x_n - x | < \epsilon_1 M = \epsilon[/itex]
So I found an [itex] n_{\epsilon_1} [/itex] such that ... [itex] | x_n - x | < \epsilon[/itex]. 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 [itex] n_{\epsilon_1} [/itex] so, by substituing , I've found [itex] n_{\frac{\epsilon}{M}} [/itex], 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