Problem 2.1.1(b)
Prove that \(1 + \pi/\sqrt{n} \rightarrow 1\) as \(n \rightarrow \infty\).
Strategy
We want to choose \(N\) such that we get this inequality (so we solve for \(n\))
$$
\begin{align*}
| 1 + \pi/\sqrt{n} - 1 | &< \epsilon \\
| \pi/\sqrt{n} | &< \epsilon \\
\pi/\sqrt{n} &< \epsilon \\
\pi/\epsilon &< \sqrt{n} \\
(\pi/\epsilon)^2 &< n.
\end{align*}
$$
Formal Proof
Let \(\epsilon > 0\). Use the Archimedean Principle to choose \(N \in \mathbb{N}\) such that
$$
\begin{align*}
N > (\pi/\epsilon)^2.
\end{align*}
$$
This implies that
$$
\begin{align*}
\epsilon > \pi/\sqrt{N}
\end{align*}
$$
When \(n \geq N\), then
$$
\begin{align*}
| (1 + \pi/\sqrt{n}) - 1| &= | \pi/\sqrt{n} | \\
&= \pi/\sqrt{n} \quad \text{(Since $n > 0$)} \\
&\leq \pi/\sqrt{N} \quad \text{(Since $n \geq N$)} \\
&< \epsilon.
\end{align*}
$$
Therefore, \(1 + \pi/\sqrt{n} \rightarrow 1\) as \(n \rightarrow \infty\). \(\ \blacksquare\)
References