# 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?

Note that $$a_{n+1} = \frac{1}{\left(a_1 + \ldots + a_{n-1}\right) + a_n} = \frac{1}{\frac{1}{a_n} + a_n} =\frac{a_n}{1 + a_n^2}$$ for $$n \geq 1$$. If you can prove that is is convergent, then, calling the limit $$\alpha$$, we must have $$\alpha = \frac{\alpha}{1+\alpha^2},$$ and solving this gives $$\alpha = 0$$.

As for proving convergence, the recurrence relation given in the form above should help you to verify that the sequence is indeed monotone decreasing, and further you should be able to check that the sequence is bounded below by zero. Then apply monotone convergence theorem.

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 $$n\to\infty$$.

1. Prove by induction that $$a_n\geq0$$ for all $$n$$ (it's pretty straightforward).

2. You'll have then $$a_{n+1}/a_n = 1/(a_n\sum_{i=1}^n a_i) < 1$$ for all $$n$$ (since $$a_1=1$$).

Thus $$a_n\geq0$$ is decreasing, converging to some limit $$x\geq0$$. Such limit must satisfy $$x=1/\sum_{n=1}^{\infty} a_n$$. If $$x>0$$, we would have convergence of $$\sum_{n=1}^{\infty}a_n$$, implying $$a_n\to 0$$, a contradiction. Therefore, $$x=\lim\limits_{n\to\infty} a_n = 0$$.