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How we can find

$$\sum_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $$

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1 Answer

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Hint: Compute $(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n+1}+\sqrt[4]{n})(\sqrt[4]{n+1}-\sqrt[4]{n})$.

  • Warm up: Simplify the expression $(a^2+b^2)(a+b)(a-b)$.
  • Pre-warm up: Simplify the expression $(a+b)(a-b)$ and deduce that $\sum\limits_{n=1}^{99}\frac1{\sqrt{n+1}+\sqrt{n}}=9$.
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And thy answer shalt be 9. – Did Sep 15 '12 at 14:09
I see what you did there ;) – Belgi Sep 15 '12 at 14:34
@Belgi Cool. Let us hope the OP too catches the hint. – Did Sep 15 '12 at 16:16
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Great. Also applies to $$\sum_{n=1}^{99999999}\frac1{(\sqrt{n+1}+\sqrt n)(\sqrt[4]{n+1}+\sqrt[4] n)(\sqrt[8]{n+1}+\sqrt[8] n)}.$$ – Hagen von Eitzen Sep 15 '12 at 16:32
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I don't think he gets the hint. What happens if you multiply top and bottom by $\sqrt[4]{n+1}-\sqrt[4]n$? – Mike Sep 15 '12 at 17:22
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