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Reciprocal Radical Sum

Problem

Consider the following series.

$$S(n) = \dfrac{1}{\sqrt{1} + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + ... + \dfrac{1}{\sqrt{n} + \sqrt{n+1}}$$

For which values of $n$ is $S(n)$ rational?


Solution

Let us consider a general term in the series and the effect of multiplying top and bottom by its conjugate, $\sqrt{k+1} - \sqrt{k}$.

$$\begin{align}\dfrac{1}{\sqrt{k} + \sqrt{k+1}} &= \dfrac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k} + \sqrt{k+1})(\sqrt{k+1} - \sqrt{k})}\\&= \dfrac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k}\sqrt{k+1} - k + (k + 1) - \sqrt{k}\sqrt{k+1}}\\&= \sqrt{k+1} - \sqrt{k}\end{align}$$

$\begin{align}\therefore S(n) &= \dfrac{1}{\sqrt{1} + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + ... + \dfrac{1}{\sqrt{n} + \sqrt{n+1}}\\&= (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + ... + (\sqrt{n+1} - \sqrt{n})\\&= \sqrt{n+1} - 1\end{align}$

Hence $S(n)$ is rational iff $n$ is one less than a perfect square; in fact, if $S(n)$ is rational then it will be integer.

Problem ID: 266 (05 Feb 2006)     Difficulty: 4 Star

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