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Assume that, for $n\ge1$,$$a_n=\sqrt{1+\left(1+\frac{1}{n}\right)^2 } +\sqrt{1+\left(1-\frac{1}{n}\right)^2 } $$

How to prove that

$$\frac{1}{a_1} +\frac{1}{a_2} +...+\frac{1}{a_{20}}$$ is an integer?

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4  
Rationalize the fractions, telescope. –  anon Aug 25 '12 at 20:25
2  
on an unrelated note, the squareroot symbol looks terrible with the concave roof. –  progressiveforest Aug 25 '12 at 20:53
    
@progressiveforest Happy now? –  Pedro Tamaroff Aug 25 '12 at 21:29
    
I wonder if behind the problem there is a fast solution ... –  Chris's sis Aug 25 '12 at 22:00
    
@peter wow ok I thought it's an inherent problem with the math typesetting here but apparently it can be improved. –  progressiveforest Aug 26 '12 at 4:58

2 Answers 2

Rationalizing the fractions yields a fairly nice form:

$$\begin{align*} \frac1{a_n}&=\frac1{\sqrt{1+\left(1+\frac{1}{n}\right)^2}+\sqrt{1+\left(1-\frac{1}{n}\right)^2}}\cdot\frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}}\\ &=\frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\left(1+\frac1n\right)^2-\left(1-\frac1n\right)^2}\\ &=\frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}}{4/n}\\ &=\frac14\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right)\;, \end{align*}$$

since $$n\sqrt{1+\left(1\pm\frac1n\right)^2}=\sqrt{n^2\left(1+\left(1\pm\frac1n\right)^2\right)}=\sqrt{n^2+\left(n\left(1\pm\frac1n\right)\right)^2}\;.$$

Thus,

$$\begin{align*} \sum_{n=1}^{20}\frac1{a_n}&=\frac14\sum_{n=1}^{20}\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right)\\ &=\frac14\left(\sum_{n=1}^{20}\sqrt{n^2+(n+1)^2}-\sum_{n=1}^{20}\sqrt{n^2+(n-1)^2}\right)\\ &=\frac14\left(\sum_{n=1}^{20}\sqrt{n^2+(n+1)^2}-\sum_{n=0}^{19}\sqrt{(n+1)^2+n^2}\right)\\ &=\frac14\left(\sqrt{20^2+21^2}+\sum_{n=1}^{19}\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n+1)^2}\right)-\sqrt{1^2-0^2}\right)\\ &=\frac14\left(\sqrt{20^2+21^2}-\sqrt1\right)\\ &=\frac14\left(\sqrt{841}-1\right)\\ &=\frac14(29-1)\\ &=7\;. \end{align*}$$

If you’re unaccustomed to summation notation, start writing out the sum longhand; you’ll find lots of telescoping.

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$$\begin{eqnarray} \frac{1}{a_n} &=& \frac{1}{\sqrt{1+\left(1+\frac{1}{n}\right)^2} + \sqrt{1+\left(1-\frac{1}{n}\right)^2}} \\ &=& \frac{1}{\sqrt{1+\left(1+\frac{1}{n}\right)^2} + \sqrt{1+\left(1-\frac{1}{n}\right)^2}} \cdot \frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2} - \sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\sqrt{1+\left(1+\frac{1}{n}\right)^2} - \sqrt{1+\left(1-\frac{1}{n}\right)^2}} \\ &=& \frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2} - \sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\left( 1+\left(1+\frac{1}{n}\right)^2 \right) - \left(1+\left(1-\frac{1}{n}\right)^2 \right)} \\ &=& \frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2} - \sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\frac{4}{n}} \\ &=& \frac{1}{4} \sqrt{n^2+\left(n+1\right)^2} - \frac{1}{4} \sqrt{n^2+\left(n-1\right)^2} \\ &=& \frac{1}{4} \sqrt{2 \left(n+\frac{1}{2}\right)^2 + \frac{1}{2}} - \frac{1}{4} \sqrt{2 \left(n-\frac{1}{2}\right)^2 + \frac{1}{2}} = f(n) - f(n-1) \end{eqnarray} $$ Now $$ \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_{20}} = f(1) - f(0) + f(2) - f(1) + \cdots + f(20) - f(19) = f(20) - f(0)= 7 $$

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