Find the limit of $\left\{x_n\right\}$ where $x_n = \frac {1}{x_1 + x_2 + \cdots + x_{n-1}}$ and $x_1 = 1$. Prove that $x_n$ converges and find the limit. Here's the question in readable form.
$$x_{1} = 1 \quad \quad x_n= \frac {1}{x_1 + x_2 + \cdots + x_{n-1}}$$
I have managed to prove that the limit exists using Monotone convergence theorem, however, i have absolutely no clue how to find the limit.
This is what I am thinking:  $$x_{n} = \frac {1}{S_{n-1}}\implies S_{n} - S_{n-1} = \frac {1}{S_{n-1}}$$
but where do i go from here?
 A: A more detailed and "down to earth" strategy is: Because $0 < x_n \le 1$ for all $n \ge 1$, you have: $x_n = \dfrac{1}{x_1+x_2+\cdots + x_{n-1}}> \dfrac{1}{n-1}\implies x_1+x_2+\cdots+x_{n-1}> 1+\dfrac{1}{2}+\cdots +\dfrac{1}{n-1}\implies \displaystyle \lim_{n\to \infty} (x_1+x_2+\cdots + x_{n-1})=+\infty\implies x_n = \dfrac{1}{x_1+x_2+\cdots + x_{n-1}}\to 0$.
A: Here's an easy way of showing divergence for the (as you noted) equivalent sequence $S_n=\frac{1}{x_n}$ which satisfies $S_n = S_{n-1}+\dfrac1{S_{n-1}}$ : $\{S_n\}$ is increasing so clearly $S_n\geq 1$ for all $n$. But then by induction, you can show that $S_n\leq n$ for all $n$: $$S_n = S_{n-1}+\dfrac1{S_{n-1}}\leq (n-1)+\dfrac1{S_{n-1}}\leq (n-1)+1=n$$
And finally, $S_n \geq H_n$, where $H_n$ is the harmonic series: $$S_n = S_{n-1}+\dfrac1{S_{n-1}} \geq H_{n-1}+\dfrac1{S_{n-1}} \geq H_{n-1}+\dfrac1n=H_n$$
A: You know the limit is nonnegative.  Suppose the limit is positive, $\lim\limits_{n\to\infty}x_n=L>0$.  As you have already noted, $x_n$ is monotone decreasing, so this implies that $x_n>L$ for all $n$, and hence $x_{n}<\dfrac{1}{(n-1)L}\to 0$.  Hence $\lim\limits_{n\to\infty}x_n=0$.
Because supposing the limit is not $0$ leads to the conclusion that the limit is $0$, the limit must be $0$.
