# Thread: determine if they are converge

1. ## determine if they are converge

$\sqrt{n+1}/2\sqrt{n}$

$(1+1/n)^{2n}$

$2^n/n^2$

$sin n / \sqrt{n}$

$\sqrt {n^2+n} - n$

can anyone tell me the general methods for simplfing these sequence so it looks more obvious?

2. Hi

$\sqrt {n^2+n} - n = \frac{\left(\sqrt {n^2+n} - n\right)\left(\sqrt {n^2+n} + n\right)}{\sqrt {n^2+n} + n}$

$\sqrt {n^2+n} - n = \frac{n}{\sqrt {n^2+n} + n}$

$\sqrt {n^2+n} - n = \frac{1}{\sqrt {1+\frac{1}{n}} + 1}$

3. Originally Posted by wopashui
$\sqrt{n+1}/2\sqrt{n}$

$\color{red}=\frac{1}{2}\sqrt{\frac{n+1}{n}}=\frac{ 1}{2}\sqrt{1+\frac{1}{n}}$

$(1+1/n)^{2n}$

$\color{red} \left(1+\frac{a}{n^k}\right)^{bn^k} \rightarrow e^{ab}$ as $\color{red} n \rightarrow \infty$

$2^n/n^2$

Use sandwich theorem ..

$sin n / \sqrt{n}$

Use sandwich theorem ..

$\sqrt {n^2+n} - n$

can anyone tell me the general methods for simplfing these sequence so it looks more obvious?
..

4. Originally Posted by General
Use sandwich theorem ..
In French we say "policemen theorem"