# Limit of a sequence of integrals involving continued fractions

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?

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

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 $$\phi=\frac{1+\sqrt 5}2$$.

• 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? – Jimmy Sabater Nov 19 at 6:44

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.

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

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