TITLE: Value of $\frac{1}{\sqrt{3}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{15}}+\cdots$ ($n$ terms)
QUESTION [11 upvotes]: Sum $$\frac{1}{\sqrt{3}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{15}}+\cdots \text{ ($n$ terms)}$$
I know how to use the telescoping series method when the terms are in product. Here i tried to make $f(n+1)-f(n)$ form but failed to do so. How do we go about with this one?
Any help is welcome,
Thanks.

REPLY [15 votes]: Multiplying top and bottom by the denominator with a $-$ sign, we can rewrite this as
$$\frac{\sqrt 7 - \sqrt 3}{7 - 3} + \frac{\sqrt{11} - \sqrt 7}{11 - 7} + \frac{\sqrt{15} - \sqrt{11}}{15 - 11} + \dots + \text{ final term}$$
This does telescope.

REPLY [6 votes]: Note that the $n^{th}$ term is
$$\dfrac1{\sqrt{4n-1} + \sqrt{4n+3}}$$
Rationalize the denominator to get
$$\dfrac1{\sqrt{4n-1} + \sqrt{4n+3}} = \dfrac{\sqrt{4n+3} - \sqrt{4n-1}}4$$
Now use telescopic cancelation to obtain what you want.