Difference between revisions of "2006 SMT/Advanced Topics Problems/Problem 5"

Latest revision as of 20:58, 27 May 2012

Problem

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

Solution

We try to rationalize the denominators. We have $\frac{1}{k\sqrt{k+2}+(k+2)\sqrt{k}}=\frac{k\sqrt{k+2}-(k+2)\sqrt{k}}{(k\sqrt{k+2})^2-((k+2)\sqrt{k})^2}=\frac{(k+2)\sqrt{k}-k\sqrt{k+2}}{2(k)(k+2)}$ .

Now, since the denominator is factorable, we try decomposing the fraction. We have $\frac{(k+2)\sqrt{k}-k\sqrt{k+2}}{2(k)(k+2)}=\frac{(k+2)\sqrt{k}}{2(k)(k+2)}-\frac{k\sqrt{k+2}}{2(k)(k+2)}=\frac{\sqrt{k}}{2k}-\frac{\sqrt{k+2}}{2(k+2)}$ .

Finally, we have $\sum_{k=1}^{\infty}\left(\frac{\sqrt{k}}{2k}-\frac{\sqrt{k+2}}{2(k+2)}\right)=\frac{1}{2}-\frac{\sqrt{3}}{6}+\frac{\sqrt{2}}{4}-\frac{2}{8}+\frac{\sqrt{3}}{6}-\frac{\sqrt{5}}{14}+\cdots$ . From here, we see that everything cancels except for the $\frac{1}{2}$ and $\frac{\sqrt{2}}{4}$ , so our final answer is $\frac{1}{2}+\frac{\sqrt{2}}{4}=\boxed{\frac{2+\sqrt{2}}{4}}$ .

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Difference between revisions of "2006 SMT/Advanced Topics Problems/Problem 5"

Latest revision as of 20:58, 27 May 2012

Problem

Solution

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