Simplify partial sums of sequence $a_n = \frac{1}{(n+1)\sqrt{n} + n\sqrt{n + 1}}$ Sum the series 
$$a_1 + a_2 + \cdots + a_{99}$$
Where  $$a_n =  \frac{1}{(n+1)\sqrt{n} + n\sqrt{n + 1}}$$
Note: $n= 1, 2, \cdots, 99$ 
Sorry for the bad formatting but I don't know how to improve it. I would greatly appreciated easy to understand working out and thanks in advance to all :)
 A: Hint
Always the same story : just multiply top and bottom by the conjugate of the denominator...
$$a_n=\frac 1{(n+1)\sqrt n + n\sqrt{n+1}} \times\frac{(n+1)\sqrt n - n\sqrt{n+1}}{(n+1)\sqrt n - n\sqrt{n+1}}$$ hence $$a_n=\frac{ (n+1)\sqrt{n}-n \sqrt{n+1}}{n (n+1)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$$
Now, see what is happening.
I am sure that you can take from here.
A: I think write following more simple
$$a_{n}=\dfrac{1}{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}$$
A: The other answers have kindly pointed out that $$a_n = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$$
There is more to the story though, because I understand you are interested in partial sums. If we consider a partial sum of this sequence:
$$\sum_{n=i}^j a_n = \bigg(\frac{1}{\sqrt{i}} - \frac{1}{\sqrt{i+1}}\bigg) + \bigg(\frac{1}{\sqrt{i+1}} - \frac{1}{\sqrt{i+2}}\bigg) + \dots + \bigg(\frac{1}{\sqrt{j}} - \frac{1}{\sqrt{j+1}}\bigg)$$
$$\sum_{n=i}^j a_n = \frac{1}{\sqrt{i}} - \bigg(\frac{1}{\sqrt{i+1}} - \frac{1}{\sqrt{i+1}}\bigg) - \frac{1}{\sqrt{i+2}} + \dots + \bigg(\frac{1}{\sqrt{j}} - \frac{1}{\sqrt{j+1}}\bigg)$$
Conveniently, the '$(n+1)$' component of one term cancels with the '$(n)$' term of the next, and the sum collapses quite nicely. (Hence why the call this a telescoping series).
$$\sum_{n=i}^j a_n = \frac{1}{\sqrt{i}} - \frac{1}{\sqrt{j+1}}$$
