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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?

Caleb Stanford
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Le Chifre
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    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? –  Mar 22 '13 at 16:54
  • yep! I mean exactly that, what you wrote below! – Le Chifre Mar 22 '13 at 17:00
  • Is the second squareroot term between each set of brackets also in the denominator? I didn't realize that... – imranfat Mar 22 '13 at 16:57

2 Answers2

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$$\sum_{n=1}^{99}\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sum_{n=1}^{99}\sqrt{n+1}-\sqrt{n}=\sqrt{100}-\sqrt{1}=9$$

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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.)

So we have,

$(1/(\sqrt1 + \sqrt2)) + (1/(\sqrt2 + \sqrt3)) + .. + (1/(\sqrt{99} + \sqrt{100}))=\displaystyle \sum_{n=1}^{99}\frac{1}{\sqrt{(n+1)}+\sqrt{n}}=\sum_{n=1}^{99}\sqrt{n+1}-\sqrt{n}=\sqrt{100}-\sqrt{1}=10-1=9$

Abhra Abir Kundu
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