<s>[INST] Approximating $\sqrt{1+\frac{1}{n}}$ by $1+\frac{1}{2n}$ I was wondering how to approximate $\sqrt{1+\frac{1}{n}}$ by $1+\frac{1}{2n}$ without using Laurent Series.
The reason why I ask was because using this approximation, we can show that the sequence $(\cos(\pi{\sqrt{n^{2}-n}})_{n=1}^{\infty}$ converges to $0$. This done using a mean-value theorem or Lipschitz (bounded derivative) argument where
$$
|\cos(\pi{\sqrt{n^{2}-n}})-\cos(\pi{n}+\pi/2)|=|\cos(\pi{\sqrt{n^{2}-n}})| 
\leq \pi|\sqrt{n^{2}-n}-n-1/2| = \pi |\frac{-1/4}{n^{2}-n+n+1/2}|
$$
I looked up  $\sqrt{1+\frac{1}{n}}$ and saw that this approximation can be obtained using Laurent series at $x=\infty$. I am not familiar with Laurent series since I have not had any complex analysis yet, but I was wondering if there was another naive way to see this?
 [/INST] Start with
$(1+x/2)^2-(1+x)
=1+x+x^2/4-(1+x)
=x^2/4
$.
Then,
$(1+x/2)^2-(1+x)
\ge 0
$,
so
$1+x/2
\ge \sqrt{1+x}
$.
Going the other way,
which is harder,
$(1+x/2)^2-x^2/4
=(1+x)
$,
so,
for $x \ge 0$,
$\begin{array}\\
\sqrt{1+x}
&=\sqrt{(1+x/2)^2-x^2/4}\\
&=(1+x/2)\sqrt{1-(x/(2(1+x/2)))^2}\\
&\ge (1+x/2)(1-(x/2(1+x/2))^2)
\quad\text{since }\sqrt{1-a}>1-a\text{ for }0 < a < 1\\
&=1+x/2-\frac{(1+x/2)x^2}{4(1+x/2)^2}\\
&=1+x/2-\frac{x^2}{4(2+x)}\\
&>1+x/2-x^2/8\\
\end{array}
$
This isn't the best,
but it is within a constant factor
of the $x^2$ term
and it is gotten by going forward,
not working backwards from
a known result.
</s>