limit of $\sqrt{12n+20}-\sqrt{12n}$ Compute the following:
$\lim\limits_{n\to \infty}\sqrt{12n+20}-\sqrt{12n}$
$\sum\limits_{n=0}^{\infty}\frac{1}{n^2+2n}$
In both I wasn't able to find anything. I tried $\lim\limits_{n\to\infty} a_n =\lim\limits_{n\to\infty} a_{n+1}$ for the first one. And I was able to show second one converge but i need the actual value.
 A: for the first one multiply and divide by $$ \sqrt {12n+20} +\sqrt {12n}$$
For the second one use partial fraction $$\frac {1}{n^2+2n} = (1/2)[(1/n) -1/{(n+2)}]$$
and you get a telescoping series. 
You should be able to handle both of them.
A: Multiply by the conjugate on top/bottom to get:
$$\frac{12n+20-12n}{\sqrt{12n+20}+\sqrt{12n}}.$$
Can you finish the rest?
A: Alternative approach for the limit:
Using that for $x>0$,$y\ge -x$
$$\sqrt{x+y}=\sqrt{\left(\sqrt x+\frac y{2\sqrt x}\right)^2-\frac{y^2}{4x}}\le\sqrt x+\frac y{2\sqrt x},$$
$$0\le\lim_{n\to\infty}(\sqrt{12n+20}-\sqrt{12n})\le\lim_{n\to\infty}\left(\sqrt{12n}+\frac{10}{\sqrt{12 n}}-\sqrt{12n}\right)=0$$
A: $$\lim\limits_{n \to \infty} \left( \sqrt{12 n + 20} - \sqrt{12 n} \right)$$
$$= 2 \lim\limits_{n \to \infty} \left( \sqrt{3 n + 5} - \sqrt{3 n} \right)$$
$$= 2 \lim\limits_{n^\prime \to \infty} \left( \sqrt{n^\prime + 5} - \sqrt{n^\prime} \right)$$
Proof:  For any $0<\epsilon$ we can find an $n$ such that the left-hand side is less than $\epsilon$:
$$\frac{5-\epsilon^2}{2 \epsilon} < \sqrt{n}$$.
Thus the limit is $0$.
