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The given problem at hand simply states,

Simplify the following expression:

$$\frac{\sqrt{10 + \sqrt{1}} + \sqrt{10 + \sqrt{2}} + \sqrt{10 + \sqrt{3}} + \dots + \sqrt{10 + \sqrt{99}}}{\sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + \sqrt{10 - \sqrt{3}} + \dots + \sqrt{10 - \sqrt{99}}}$$

Notice that the terms in the denominator are the conjugates of the terms in the numerator and that the whole expression can be rewritten as a quotient of two sums in summation notation:

$$\frac{\sum_{i = 1}^{99}{\sqrt{10 + \sqrt{i}}}}{\sum_{j = 1}^{99}{\sqrt{10 - \sqrt{j}}}}$$

But this is all that I have gathered. I don't know what to do. Any hints?

J C
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1 Answers1

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Since the question has already an answer, this is just for the fun of it.

Replacing the summation by the integral and then considering $$F(a)=\frac{\int_1^{a^2-1}\sqrt{a+\sqrt x}\,\,dx }{\int_1^{a^2-1}\sqrt{a-\sqrt x}\,\,dx }$$

$$\int\sqrt{a+\sqrt x}\,\,dx =\frac{4}{15} \sqrt{a+\sqrt{x}} \,\left(3x+a \sqrt{x}-2a^2\right)$$ $$\int\sqrt{a-\sqrt x}\,\,dx=\frac{4}{15} \sqrt{a-\sqrt{x}}\, \left(3 x-a \sqrt{x}-2 a^2\right)$$

Assuming $a>\sqrt 2$ we find $$F(a)=\frac{\sqrt{a+1} \left(2 a^2-a-3\right)+\sqrt{a+\sqrt{a^2-1}} \left(a^2+a\sqrt{a^2-1}-3\right)}{\sqrt{a-1} \left(2 a^2+a-3\right)+\sqrt{a-\sqrt{a^2-1}} \left(a^2-a\sqrt{a^2-1} -3\right)}$$

Now, a tedious series of simplifications leads to $$\color{red}{F(a)=1+\sqrt2 \qquad \forall a}$$

Claude Leibovici
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