Ftyjtk

The following question was asked in a calculus exam in UNI, a Peruvian university. It is meant to be for freshman calculus students.

Find $lim_{n to infty} A_n $ if

$$ A_1 = intlimits_0^1 frac{dx}{1 + sqrt{x} }, ; ; ; A_2 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+sqrt{x}} }, ; ; ; A_3 =
intlimits_0^1 frac{dx}{1 + frac{1}{1+frac{1}{1+sqrt{x}}} },
...$$

First of all, I think this is a hard question for a midterm exam, but anyway, notice that we can calculate $A_1$ by making $t=sqrt{x}$

$$ A_1 = intlimits_0^1 frac{2 t dt }{1+t} = 2 intlimits_0^1 dt - 2 intlimits_0^1 frac{dt}{1+t}=2-2(ln2)=2-ln2^2 $$

Now, as for $A_2$ I would do $t = frac{1}{1+sqrt{x}}$ which gives $d t = frac{ dx}{2 sqrt{x} (1+sqrt{x})^2} = frac{t^2 dx}{2 (t-1)}$ thus upon sustituticion we get

$$ A_2 = - intlimits_1^{1/2} frac{2 (t-1) }{t^2(1+t) } dt $$

which can easily solved by partial fractions or so. But, apparently this is not the way this problem is meant to be solved as this exam contained 4 questions to be solved in an hour. What is the trick, if any, that can be used to solve this problem without doing the unellegant partial fractions?

By the substitution $t=sqrt x$,
$$A_n=int^1_0f_n(t^2)(2tdt)$$

$f_n(t^2)$ is of the form
$$f_n(t^2)=frac{a_n+b_nt}{c_n+d_nt}$$

We have the recurrence relation
$$a_{n+1}=c_n$$
$$b_{n+1}=d_n$$
$$c_{n+1}=a_n+c_n$$
$$d_{n+1}=b_n+d_n$$

Or
$$c_{n+1}=c_n+c_{n-1}$$
$$d_{n+1}=d_n+d_{n-1}$$

which are the Fibonacci recurrence with initial conditions
$$c_0=1, c_1=1$$
$$d_0=0, d_1=1$$

I think you can now proceed.

Also, the general term of Fibonacci sequence $0,1,1,cdots$ is
$$frac{phi^n-overlinephi^n}{sqrt5}$$ where $phi=frac{1+sqrt 5}2$.

If I were taking that exam, I'd speculate covergence and write the integrand for $A_infty$ as
$$ S_infty(x) = frac{1}{ 1 + frac{1}{1+frac{1}{1+frac{1}{ddots} }}} = frac{1}{1+S_infty(x)}$$
Solve the resulting quadratic for $S_infty^2(x) + S_infty(x) -1 = 0$ for $S_infty(x)=frac{-1+sqrt{5}}{2}$. Then we immediately have $A_infty = S_infty$.

Then, I'd sit there and wonder what they intended for me to actually show on a freshman calculus exam.

Let $A_n = intlimits_0^1 f_n(x) dx$. First show that each $f_n$ is a monotonically increasing or decreasing function in the range $[0,1]$, and hence $A_n$ lies between $f_n(0)$ and $f_n(1)$. Then note that

$f_n(0) = 1, frac{1}{2}, frac{2}{3}, frac{3}{5}, dots$

are the convergents of the continued fraction expansion of $phi$, and $f_n(1) = f_{n+1}(0)$. So we have

$frac{1}{2} le A_1 le 1$

$frac{1}{2} le A_2 le frac{2}{3}$

$frac{3}{5} le A_3 le frac{2}{3}$

and so on.

So the sequence $A_n$ is squeezed between consecutive convergents of the continued fraction expansion of $phi$. And so $lim_{n to infty} A_n = phi$.