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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?

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    $\begingroup$ 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? $\endgroup$
    – user14972
    Mar 22, 2013 at 16:54
  • $\begingroup$ Is the second squareroot term between each set of brackets also in the denominator? I didn't realize that... $\endgroup$
    – imranfat
    Mar 22, 2013 at 16:57
  • $\begingroup$ yep! I mean exactly that, what you wrote below! $\endgroup$
    – Le Chifre
    Mar 22, 2013 at 17:00

2 Answers 2

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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,

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

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  • $\begingroup$ Do the parentheses indicate $\frac{1}{\sqrt{n+1}+\sqrt n}$ or $\frac{1}{\sqrt n}+\sqrt{n+1}$? $\endgroup$
    – user47805
    Mar 22, 2013 at 16:55
  • $\begingroup$ Now its clear @user47805, i guess $\endgroup$
    – Abhra Abir Kundu
    Mar 22, 2013 at 16:56
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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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