Dtyjlui

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?

As requested by the OP in a comment deleted by a moderator, switching of limit and integral sign is avoided as it requires a higher-than-expected level of knowledge for justification. Thus, a ‘simpler’ approach is presented.

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 $lim_{n to infty} A_n = phi$.

$sqrt{x}$
is a McGuffin.

More generally,
let
$f_1 = frac{1}{1 + g(x) }
$
where
$g'(x) > 0,
g(0) = 0
$,
$f_n(x)
=frac{1}{1+f_{n-1}(x)}
$,
and
$A_n = int_0^1 f_n(x) dx
$.

Then
$f_n(x)
to dfrac{sqrt{5}-1}{2}
$.

Note:
I doubt that any of this
is original,
but this was all done
just now by me.

Proof.

$begin{array}\
f_n(x)
&=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
&=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
&=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
end{array}
$

Therefore,
if $f_{n-2}(x) > 0$
then
$frac12 < f_n(x)
lt 1$.

Similarly,
if $f_{n-1}(x) > 0$
then
$0 < f_n(x)
lt 1$.

$begin{array}\
f_n(x)-f_{n-2}(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
&=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
end{array}
$

$begin{array}\
f_n(x)+f_n^2(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
&=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
text{so}\
1-f_n(x)-f_n^2(x)
&=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
end{array}
$

Therefore
$1-f_n(x)-f_n^2(x)$
has the same sign as
$1-f_{n-2}(x)-f_{n-2}^2(x)$.
Also,
$|1-f_n(x)-f_n^2(x)|
lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
$
so
$|1-f_n(x)-f_n^2(x)|
to 0$.

Let
$p(x) = 1-x-x^2$
and
$x_0 = frac{sqrt{5}-1}{2}
$
so
$p(x_0) = 0$,
$p'(x) < 0$ for $x ge 0$.

Since
$f_n(x) > 0$,
$f_n(x)
to x_0$.