limits - Prove convergence of sequence given by $a_{1}=1$ and $a_{n+1}= frac{1}{a_1+a_2+ldots+a_n}$

For sequence given by $a_{1}=1$ and $a_{n+1}= \frac{1}{a_1+a_2+\ldots+a_n}$ I have to prove that it converges to some number and find this number.

I tried toshow that it's monotonic by calculating
$$
\frac{a_{n+1}}{a_{n}} = \frac{1}{a_{n}(a_1+a_2+\ldots+a_n)}
$$
but I cannot say anything about the denominator. How can I try to find it's limit?

Answer

Let $s_n = \sum\limits_{k=1}^n a_k$. We can rewrite the recurrence relation as

$$s_{n+1} - s_n = a_{n+1} = \frac{1}{s_n} \quad\implies s_{n+1} = s_n + \frac{1}{s_n}$$

This implies
$$s_{n+1}^2 = s_n^2 + 2 + \frac{1}{s_n^2} \ge s_n^2 + 2$$

So for all $n > 1$, we have

$$s_n^2 = s_1^2 + \sum_{k=1}^{n-1} (s_{k+1}^2 - s_k^2) \ge 1 + \sum_{k=1}^{n-1} 2 = 2n - 1$$

Since all $a_n$ is clearly positive, we have $\displaystyle\;0 < a_n = \frac{1}{s_{n-1}} \le \frac{1}{\sqrt{2n-3}}$.

By squeezing, $a_n$ converges to $0$ as $n\to\infty$.