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?
 A: For any $x>0$, $\sqrt{1+x}\leq 1+\frac{x}{2}$ is trivial by squaring. On the other hand:
$$ 1+\frac{x}{2}-\sqrt{1+x} = \frac{\frac{x^2}{4}}{1+\frac{x}{2}+\sqrt{1+x}}\leq\frac{x^2}{8+2x} $$
gives:

$$ 1+\frac{x}{2}-\frac{x^2}{8+2x}\leq \sqrt{1+x} \leq 1+\frac{x}{2}-\frac{x^2}{8+4x}.$$

A: This requires nothing more than just the definition of a derivative. The function $f(x)=\sqrt{1+x}$ has derivative $f'(x)=\frac{1}{2\sqrt{1+x}}$ for $x\geq 0$. By the fundamental definition of derivatives:
$$f(x)-f(0)=f'(0)x+\epsilon(x),$$
where $\lim_{x\rightarrow 0^+}\epsilon(x)/x=0$ and  $\lim_{x\rightarrow 0^+} \epsilon(x)=0$. If you're unfamiliar with this fact, divide both sides by $x$ and take the limit, invoking the definition of the derivative on the left side. To avoid confusion, it's actually that $\epsilon(x):=f(x)-f(0)-f'(0)$, which is a defines $\epsilon(x)$. 
It follows that:
$$f(x)=1+\frac{x}{2}+\epsilon(x).$$
Now plug in $x=1/n$ and observe that $\epsilon(1/n)/(1/n)$ is going to zero, which means that $\epsilon(1/n)$ must be considerably smaller than $(1/n)$ as $n$ gets larger. It follows that:
$$\sqrt{1+1/n}=1+2/n+(\mbox{something much smaller than }1/n),$$
an approximation which will tend to get better as $n$ gets larger.
A: Try
$\sqrt{1+x} \approx 1 + \alpha x + \beta x^2 + O(x^3)$
$1 + x = 1 + 2 \alpha x + (\alpha^2 + 2 \beta)x^2 + O(x^3)$
$\alpha = \frac 12$, $\quad \beta = -\frac 18$
A: This is not so different from other approaches (exactly equivalent to some), but has a different perspective based on pure computation. 
Suppose $a$ is an estimate for the square root of $b$ so that $$a^2=b+h$$where $h$ is taken to be small (see below).
Then $$\left(a-\frac h{2a}\right)^2=a^2-h+\left(\frac {h^2}{4a^2}\right)=b+\left(\frac {h^2}{4a^2}\right)$$
And this gives a better estimate provided $|h|\gt\cfrac {h^2}{4a^2}$ or $|h|\lt 4a^2$ i.e. $h$ is small (definition).
If $h$ is small then the process can be repeated, and is clearly seen to be convergent (the error reduces faster than by a constant ratio).
This does not work, obviously, for $a=0$. For small $a$ it can sometimes help computationally to compute the square root of $\frac 1a$.
A: 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.
