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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

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#1 h Mar 20, 2010, 7:29 PM • 4 Y

Y by Adventure10, Mango247, At777, and 1 other user

Evaluate: $\sum\limits_{k=1}^\infty \frac{1}{k\sqrt{k+2}+(k+2)\sqrt{k}}$

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#2 h Mar 21, 2010, 11:34 AM • 4 Y

Y by AlastorMoody, Adventure10, Mango247, and 1 other user

$\sum\limits_{k=1}^\infty\frac{1}{k\sqrt{k+2}+(k+2)\sqrt{k}} =\\ \sum\limits_{k=1}^\infty\frac{(k+2)\sqrt{k}-k\sqrt{k+2}}{(k\sqrt{k+2}+(k+2)\sqrt{k})((k+2)\sqrt{k}-k\sqrt{k+2})} = \\ \sum\limits_{k=1}^\infty\frac{(k+2)\sqrt{k}-k\sqrt{k+2}}{2k(k+2)} =\\ \frac{1}{2}\sum\limits_{k=1}^\infty\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+2}} =\\ \frac{1}{2}(1 + \frac{1}{\sqrt{2}})$

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#3 h Apr 22, 2025, 12:18 AM

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Let $T_k$ denote the general term of the given sum $S$ .

Then, $T_k = \frac{1}{k\sqrt{k + 2} + (k + 2)\sqrt{k}} = \frac{1}{\sqrt{k(k + 2)} (\sqrt{k} + \sqrt{k + 2})} = \frac{\sqrt{k + 2} - \sqrt{k}}{2\sqrt{k(k + 2)}} = \frac{1}{2}\left(\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 2}}\right)$
Now, $S = \sum_{k=1}^\infty T_k = \frac{1}{2}\left(1 + \frac{1}{\sqrt{2}}\right)$
Therefore, $S = \boxed{\frac{\sqrt2 + 1}{2\sqrt2}}$

This post has been edited 11 times. Last edited by P162008, Apr 22, 2025, 6:46 PM
Reason: Typo

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#5 h Apr 22, 2025, 6:12 PM

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