# How to effectively calculate $(1/\sqrt1 + \sqrt2) + (1/\sqrt2 + \sqrt3) +\cdots + (1/\sqrt{99} + \sqrt{100})$ [duplicate]

I have this series:

$$\frac{1}{\sqrt1 + \sqrt2} +\frac{1}{\sqrt2 + \sqrt3} +\frac{1}{\sqrt3 + \sqrt4} +\cdots+\frac{1}{\sqrt{99} + \sqrt{100}}$$

My question is, what approach would you use to calculate this problem effectively?

• You wrote $$\left( \frac{1}{\sqrt{1}} + \sqrt{2} \right) + \left( \frac{1}{\sqrt{2}} + \sqrt{3} \right) + \cdots$$ did you mean $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots$$ instead?
– user14972
Mar 22, 2013 at 16:54
• Is the second squareroot term between each set of brackets also in the denominator? I didn't realize that... Mar 22, 2013 at 16:57
• yep! I mean exactly that, what you wrote below! Mar 22, 2013 at 17:00

Hint :$\displaystyle \frac{1}{\sqrt{(n+1)}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$(By multiplying the numerator and the denominator by multiplying $(\sqrt{n+1}-\sqrt{n})$ to both numerator and denominator.)
$(1/(\sqrt1 + \sqrt2)) + (1/(\sqrt2 + \sqrt3)) + .. + (1/(\sqrt{99} + \sqrt{100}))=\displaystyle \sum_{n=1}^{99}\frac{1}{\sqrt{(n+1)}+\sqrt{n}}=\sum_{n=1}^{99}\sqrt{n+1}-\sqrt{n}=\sqrt{100}-\sqrt{1}=10-1=9$
• Do the parentheses indicate $\frac{1}{\sqrt{n+1}+\sqrt n}$ or $\frac{1}{\sqrt n}+\sqrt{n+1}$? Mar 22, 2013 at 16:55
$$\sum_{n=1}^{99}\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sum_{n=1}^{99}\sqrt{n+1}-\sqrt{n}=\sqrt{100}-\sqrt{1}=9$$