Value of $\sum \limits_{k=1}^{81} \frac{1}{\sqrt{k} + \sqrt{k+1}} = \frac{1}{\sqrt{1} + \sqrt{2}} + \cdots + \frac{1}{\sqrt{80} + \sqrt{81}}$?

I tried my best, but I am totally clueless about it. Worse thing is we were supposed to arrive at the answer in approximately $2$ minutes. The correct answer is $8$, right? Can you kindly explain how to arrive at it? I hope it won’t be too much bother. Thank you.

• Are you sure you've got the question right? If the terms were multiplied, it would be easy. Feb 22 '13 at 6:39
• @JanDvorak Sorry for the delay in replying to you.Yes, I have got the question right.I double-check things before putting them up on this reputed maths forum. Feb 22 '13 at 12:18

Note that

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

Then you have a telescoping sum

• Thanks Muzzlator for your fine answer. Feb 22 '13 at 12:19

As \begin{align} \frac{1}{\sqrt{i}+\sqrt{i+1}} = \frac{\sqrt{i}-\sqrt{i+1}}{\sqrt{i}-\sqrt{i+1}} \cdot \frac{1}{\sqrt{i}+\sqrt{i+1}} = \frac{\sqrt{i}-\sqrt{i+1}}{-1} = {\sqrt{i+1}-\sqrt{i}} \end{align} Thus \begin{align} \sum_{i=1}^{80} \frac{1}{\sqrt{i}+\sqrt{i+1}} = \sum_{i=1}^{80} {\sqrt{i+1}-\sqrt{i}} = \sqrt{80+1} -\sqrt{1} = 8 \end{align}

• A prof of mine always said, that "text is the shortest connection between two formulas" ;) Feb 22 '13 at 6:51

$$\frac{\sqrt{2} - \sqrt{1} }{(2-1)} +\ldots +\frac{\sqrt{81} - \sqrt{80} }{(81-80)}=\sqrt{81}-\sqrt{1}=9-1=8$$