Question

7B1%7D%7B2%7D%20%5Csqrt%7Bx%20%2B%20%5Cfrac%7B1%7D%7B4%7D%20%5Csqrt%7Bx%20%2B%20%5Cfrac%7B1%7D%7B8%7D%20%5Csqrt%7Bx%20%2B%20%5Cfrac%7B1%7D%7B16%7D%20%5Cdots%7D%20%7D%20%7D%20%7D%20%5C%3A%20%2B%20%5Csqrt%7Bx%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7Bx%20-%20%5Cfrac%7B1%7D%7B4%7D%20%5Csqrt%7Bx%20-%20%5Cfrac%7B1%7D%7B8%7D%20%5Csqrt%7Bx%20-%20%5Cfrac%7B1%7D%7B16%7D%20%5Cdots%7D%20%7D%20%7D%20%7D%20%5Cright%29%20dx%20%5C%5C%20" id="TexFormula1" title=" \tiny\int_{e}^{{e}^{2}} \left( \sqrt{x + \frac{1}{2} \sqrt{x + \frac{1}{4} \sqrt{x + \frac{1}{8} \sqrt{x + \frac{1}{16} \dots} } } } \: + \sqrt{x - \frac{1}{2} \sqrt{x - \frac{1}{4} \sqrt{x - \frac{1}{8} \sqrt{x - \frac{1}{16} \dots} } } } \right) dx \\ " alt=" \tiny\int_{e}^{{e}^{2}} \left( \sqrt{x + \frac{1}{2} \sqrt{x + \frac{1}{4} \sqrt{x + \frac{1}{8} \sqrt{x + \frac{1}{16} \dots} } } } \: + \sqrt{x - \frac{1}{2} \sqrt{x - \frac{1}{4} \sqrt{x - \frac{1}{8} \sqrt{x - \frac{1}{16} \dots} } } } \right) dx \\ " align="absmiddle" class="latex-formula">

Answer 1

We have the identity

$\left(\sqrt x + \dfrac1{2^n}\right)^2 = x + \dfrac1{2^{n-1}} \sqrt x + \dfrac1{2^{2n}}$

Take the square root of both sides and rearrange terms on the right to get

$\sqrt x + \dfrac1{2^n} = \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt{x} + \dfrac1{2^{n+1}}\right)}$

Decrementing n gives

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \left(\sqrt{x} + \dfrac1{2^{n}}\right)}$

and substituting the previous expression into this, we have

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right) } }$

Continuing in this fashion, after k steps we would have

$\sqrt x + \dfrac1{2^{n-k}} = \sqrt{x + \dfrac1{2^{n-(k+1)}} \sqrt{x + \dfrac1{2^{n-k}} \sqrt{x + \dfrac1{2^{n-(k-1)}} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

After a total of n - 2 steps, we arrive at

$\sqrt x + \dfrac14 = \sqrt{x + \dfrac12 \sqrt{x + \dfrac1{2^2} \sqrt{x + \dfrac1{2^3} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

Then as n goes to infinity, the first nested radical converges to √x + 1/4. Similar reasoning can be used to show the other nested radical converges to √x - 1/4. Then the integral reduces to

$\displaystyle \int_e^{e^2} \left(\sqrt x - \frac14\right) + \left(\sqrt x + \frac14\right) \, dx = 2 \int_e^{e^2} \sqrt x \, dx = \boxed{\frac43 \left(e^3 - e^{\frac32}\right)}$