Using Negation of Convergence Def to Disprove

Question

I want to show that the sequence x defined by

x = 1/n

does not converge to 1 by using the negation of the definition of convergence. The negation of convergence is "There exists an ε for all N∈ℕ such that there exists an n≥N and |x - L| ≥ ε."

I have proven that x= 1/n converges to 0 before using the original definition but I am struggling to find a way to use the negation to prove that it does NOT converge to 1.

Answer 1

If you meant with your statement about the negation of the convergence the equivalent to

A sequence $ (x_k)_{k \in \mathbb{N}} $ does not converge against L if and only if for all $ N \in \mathbb{N} $ there exists an $ \epsilon > 0 $ such that there exists an $ n \geq N $ such that $ |x_n - L| \geq \epsilon $

, then you have made a mistake, since e.g. for any $ N \in \mathbb{N}, \epsilon = \frac{1}{N}: |\frac{1}{N} - 0| = \frac{1}{N} \geq \epsilon $, although the sequence $ (\frac{1}{k})_{k \in \mathbb{N}} $ converges against 0.

A correct negation of the convergence would be

A sequence $ (x_k)_{k \in \mathbb{N}} $ does not converge against L if and only if there exists an $ \epsilon > 0 $ such that there does not exist an $ N \in \mathbb{N} $ such that $ |x_n - L| < \epsilon $ for all n $ \geq $ N

which holds for $ (\frac{1}{k})_{k \in \mathbb{N}} $ as for for all $ N \in \mathbb{N}, N \geq 2 $, $ \epsilon = 0.5 $, it is the case for all n $ \geq N $ that $ |x_n - 1| >= |x_2 - 1| = |0.5 - 1| = 0.5 >= 0.5 = \epsilon $. To be strict, you would also have to show that the statement highlighted above holds even for N = 1, but I am sure you see why it does :)