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.
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.
$$\lim\limits_{n \to \infty} \left( \sqrt{12 n + 20} - \sqrt{12 n} \right)$$

$$= 2 \lim\limits_{n \to \infty} \left( \sqrt{3 n + 5} - \sqrt{3 n} \right)$$

$$= 2 \lim\limits_{n^\prime \to \infty} \left( \sqrt{n^\prime + 5} - \sqrt{n^\prime} \right)$$

Proof:  For any $0&lt;\epsilon$ we can find an $n$ such that the left-hand side is less than $\epsilon$:

$$\frac{5-\epsilon^2}{2 \epsilon} &lt; \sqrt{n}$$.

Thus the limit is $0$.