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

$$ A_1 = \int\limits_0^1 \frac{dx}{1 + \sqrt{x} }, \; \; \; A_2 = \int\limits_0^1 \frac{dx}{1 + \frac{1}{1+\sqrt{x}} }, \; \; \; A_3 = \int\limits_0^1 \frac{dx}{1 + \frac{1}{1+\frac{1}{1+\sqrt{x}}} }, ...$$

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 $t=\sqrt{x}$

$$ A_1 = \int\limits_0^1 \frac{2 t dt }{1+t} = 2 \int\limits_0^1 dt - 2 \int\limits_0^1 \frac{dt}{1+t}=2-2(\ln2)=2-\ln2^2 $$

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

$$ A_2 = - \int\limits_1^{1/2} \frac{2 (t-1) }{t^2(1+t) } dt $$

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?

  • $\begingroup$ Can you prove it is valid to change the position of $\lim$ and $\int$? $\endgroup$ Nov 19, 2018 at 5:59
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    $\begingroup$ @Szeto: Could you post your comment as an answer (with details)? ( ̄▽ ̄) $\endgroup$
    – Tianlalu
    Nov 19, 2018 at 6:17
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    $\begingroup$ I think this is meant to be solved without that dominated convergence theorem as it is meant for first-year students. $\endgroup$
    – James
    Nov 19, 2018 at 6:17
  • 2
    $\begingroup$ @JimmySabater Although first-year students have never heard of the dominated convergence theorem, they often act as though it were obvious. $\endgroup$ Nov 19, 2018 at 16:35

4 Answers 4


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.

  • $\begingroup$ @zahbaz Is there a particular reason you choose the positive root to the quadratic equation? $\endgroup$ Nov 19, 2018 at 21:12
  • $\begingroup$ @JonathanChiang Every term in the continued fraction $S_\infty(x)$ is positive and all operations involve addition, so it must be positive. $\endgroup$
    – zahbaz
    Nov 19, 2018 at 21:27
  • $\begingroup$ @zahbaz tell that to $\zeta(-1)$ $\endgroup$
    – Peter
    Nov 20, 2018 at 12:57

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 $$\frac{\phi^n-\overline\phi^n}{\sqrt5}$$ where $\phi=\frac{1+\sqrt 5}2$.

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    $\begingroup$ It is very nice, but I believe this is a hard question to ask in a midterm for beginners. I dont think there is a more elementary solution, correct? $\endgroup$
    – James
    Nov 19, 2018 at 6:44

Let $A_n = \int\limits_0^1 f_n(x) dx$. 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.


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

  • $\begingroup$ what do you mean by Mcguffin? $\endgroup$
    – James
    Dec 28, 2018 at 8:52
  • $\begingroup$ Any g(x) > 0 will do. MacGuffin g.co/kgs/wDd8KQ $\endgroup$ Dec 28, 2018 at 14:26

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