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If $y > 0$, show that there exists $n \in \mathbb{N}$ such that $1/2^n < y$.

The book hints that since $n < 2^n$, it follows that $1/n > 1/2^n$.

So, I start establishing that $n < 2^n$ for all $n \in \mathbb{N}$.

Proof by induction:

Base step: $n = 1$.

When $n = 1$, we have $1 < 2^1$, which is true. Thus, $n < 2^n$ when $n = 1$.

Induction step:

Assume that $n < 2^n$ is true up to $n$. Need to show that the statement is true for $n + 1$, or that $n + 1 < 2^{(n + 1)}$ is true.

$n + 1 < 2^n*2^1 = 2^n*2$

Then, by the induction hypothesis, since $n < 2^n$, it follows that $2 * n < 2 * 2^n$ or $n + n < 2 * 2^n$. And since $n > 1$, it follows that $n + n > n + 1$. Consequently, $n + 1 < n + n < 2 * 2^n$, and by the transitive property, $n + 1 < 2 * 2^n$. Thus, the statement is true for $n + 1$, and $n < 2^n$ for all $n \in \mathbb{N}$.

Now that we established that $n < 2^n$, we know that $1/n > 1/2^n$. So, for any $y > 0$, pick $n \in \mathbb{N}$ so that $1/n < y$. This will give us $1/2^n < 1/n < y$, and by the transitivity property $1/2^n < y$.

What's wrong with this proof?

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1… So just take $1/n<y$ and you're done. – Jose Antonio Feb 18 '14 at 1:47
I'm a bit confused with the hint given by the book because $\frac{1}{2^n} \longrightarrow 0$ when $n \rightarrow +\infty$ so we know that $\forall y > 0$ there exist $n \in \mathbb{N}$ such that $\frac{1}{2^n} < y$... – CharlesPerry Feb 18 '14 at 1:48
This is overkill. The statement is obvious. It is equivalent to choosing $n>-\log(\epsilon)/\log(2)$. Now this is well-defined and since we can cover $\mathbb{R}$ by the union of all sets $[k-1,k), k \in \mathbb{Z}$, we have that $-\log(\epsilon)/\log(2) \in [k-1,k)$ for some $n \in \mathbb{Z}$. So choose $n+1$ and we are done. But this is overkill. – Chris K Feb 18 '14 at 3:08
I'm still at the baby Real Analysis. – user129360 Feb 18 '14 at 3:21

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