Let $δ$ $∈$ $\mathbb{R}$ , Prove that there exists an $n$ ∈ $\mathbb{N}$ such that $\frac {2}{n + 2}$ $<$ $δ$.

Question

Question: Let $δ$ $∈$ $\mathbb{R}$ and $δ$ $>$ $0$. Prove that there exists an $n$ ∈ $\mathbb{N}$ such that $\frac {2}{n + 2}$$<$ $δ$.

I get that there is an n that satisfies this, but seemingly only under certain $δ$.

How do I go about proving this formally?

I rearranged the inequality to $n$ $<$ $2$ ($\frac 1δ$ - 1) so $n$ $\ge$ $1$ and for 2((1/δ)-1) > $0$,

$δ$ must be $0<δ<1$. Any advice on how to continue or other approaches would be appreciated.

Answer 1

Based on your work, you can choose $n$ to be any natural number for which

$$n \ge 2 \left(\frac 1 {\delta} - 1\right)$$

and the original inequality holds. For example, if $\delta < 1$ then

$$n = \left\lceil 2 \left(\frac 1 {\delta} - 1\right)\right\rceil + 42$$ works.

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If $\delta \ge 1$, then literally any selection of $n$ works, since the given fraction is always at most $1$.

Answer 2

Hint:

If $\lim_{n \rightarrow \infty}\frac{2}{n+2} = 0$, then for all $\delta >0$ there exists $N \in \mathbb{N}$ such that $|\frac{2}{n+2}| < \delta$ whenever $n>N$.

Answer 3

You're almost there, but not quite. Since it is a proof about some $n$ $\in$ $\mathbb{N}$ existing you must only provide a formula for choosing this $n$. Based on your work we have $n$ $>$ $2 \over \delta$ $- 2$. Using this formula, just choose $n$ to be any natural number that satisfies this condition.