# Proving that a recursive converges and calculate the limit?

I've been working on this problem for the past few hours, and I've made some progress, but I'm stuck.

Say we have some sequence $x_1 = 1$ and $x_{n} = \frac{1}{x_1 + ... + x_{n-1}}, n \geq 1$. I need to prove convergence and obtain the limit.

I was able to prove by induction on $n$, that $x_n = \frac{1}{x_{n-1} + \frac{1}{x_{n-1}}}, n \geq 3$

I was thinking that I could use monotone convergence theorem, but I'm not sure how to apply this to a recursive sequence that's similar to a continued fraction. My thinking (completely non-rigorously) is to prove that the denominator $x_{n-1} + \frac{1}{x_{n-1}}$increases monotonically, and since $x_1= 1$, it's bounded above by $1$. But I can't figure out how to prove this (maybe using the $\epsilon$ definition) and compute a limit mathematically. I hypothesize the limit is $\frac{1}{4}$, but this was just repeated calculations.

The alternative is to prove it's Cauchy, but I don't see how I can choose $\epsilon$ so that it does not depend on $n$.

So if $x_n$ is a bounded monotone sequence, then we are done.
$x_n$ is bounded. To see this, see that from the definition of $x_n$, we have $x_n > 0$ for each $n$. Furthermore, for $n > 1$, $x_n = \dfrac{1}{x_{n-1} + \frac{1}{x_{n-1}}} = \dfrac{1}{\left(\sqrt {x_{n-1}} - \frac{1}{\sqrt{x_{n-1}}}\right)^2 + 2} \leq \dfrac 12$.
Furthermore, we know $\frac{1}{x_n} =x_{n-1} + \frac{1}{x_{n-1}} \geq \frac{1}{x_{n-1}}$, so that $x_{n-1} \geq x_n$. Therefore, $x_n$ is a bounded monotone sequence, and therefore has a limit(monotone convergence theorem, as you call it).
Now, the limit, if iterated under the recurrence, must be itself. That is, if $x$ is the limit, then $x =\frac{x}{x^2 + 1}$, which actually gives $x = 0$. You can see that if $\frac 14$ were to be a limit, then it would have to satisfy this equation, but it does not.
Set $a_n=x_1+...+x_n$ then $$a_{n+1}-a_n=\frac1{a_n}.$$ In the loosely related differential equation $\dot u=\frac1u$ you would look at solutions $u^2=2t+C$, thus consider the square of the sequence $(a_n)$ to find $$a_{n+1}^2=a_n^2+2+\frac1{a_n^2}$$ so that with $a_1=1$ you get $$a_n^2\ge 2n-1$$ so that for $n\ge 2$ $$x_n=\frac1{a_n-1}\le \frac1{\sqrt{2n-3}}.$$ This upper bound clearly converges to zero.