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How we can find

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

2 Answers 2

23

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 9.2012-09-15
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    I see what you did there ;)2012-09-15
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    @Belgi Cool. Let us hope the OP too catches the hint.2012-09-15
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    Oh.Little more please .2012-09-15
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    Great. Also applies to $$\sum_{n=1}^{99999999}\frac1{(\sqrt{n+1}+\sqrt n)(\sqrt[4]{n+1}+\sqrt[4] n)(\sqrt[8]{n+1}+\sqrt[8] n)}.$$2012-09-15
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    @HagenvonEitzen Indeed. Good hint, I like it. brab: was this *little more* by Hagen (plus my Edit) enough to make you see the light?2012-09-15
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    I don't think he gets the hint. What happens if you multiply top and bottom by $\sqrt[4]{n+1}-\sqrt[4]n$?2012-09-15
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    @Mike Thanks for the help.2012-09-16
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$\frac {1}{(\sqrt {n+1}+\sqrt {n})(\sqrt[4] {n+1}+\sqrt[4] {n})}=\frac {(\sqrt[4] {n+1}-\sqrt[4] {n})}{(\sqrt {n+1}+\sqrt {n})(\sqrt[4] {n+1}+\sqrt[4] {n})(\sqrt[4] {n+1}-\sqrt [4]{n})}=\frac {(\sqrt[4] {n+1}-\sqrt [4]{n})}{n+1-n}=\sqrt[4] {n+1}-\sqrt[4] {n}$ so the sum is transformed to $\sum _{n=1}^{9999} \sqrt[4] {n+1}-\sqrt[4] {n}=\sqrt[4] {10000}-\sqrt[4] {1}=9$.