# $1 + \frac1{2+\frac1{2}} + \dots$ diverges.

Let $$\{ s_n\}$$ be an infinite sequence with $$s_1 = 1$$ $$s_2=\frac1{2+\frac12}$$ $$s_3 = \frac{1}{3+\frac{1}{3+\frac13}}$$ $$\vdots$$ I would like to show $$\displaystyle\sum_{i=1}^\infty s_i$$ diverges. My attempt:

Note that $$\lim_{k \to \infty}[ 0; k_1,k_2 \dots k_j] = 0$$ for all $$j>0$$ (where $$j$$ denotes the $$j$$-th $$k$$) with $$[c_0; c_1,c_2 \dots c_j]$$ being continued fraction notation. Then as $$n \to \infty$$ $$s_n = \frac1{n+\frac1{n+\frac1{\ddots}}} \to 0$$ Now, I assert $$s_n \to \frac{1}{n+s_n}$$ as $$n \to \infty$$. This implies $$s_n \to \frac1n$$ Now let $$\{ a_m\}$$ be an infinite sequence with $$a_q = \frac1q$$. Given that $$\sum_{i=1}^\infty a_i$$ diverges, $$s_n \to a_n$$, and $$a_n,s_n$$ are both always positive, it must be the case that $$\sum_{i=1}^\infty s_i$$ diverges. In other words, $$1+\frac{1}{2+\frac{1}{2}}+\frac1{3+\frac1{3+\frac1{3}}} + \frac1{4+\frac1{4+\frac1{4+\frac1{4}}}} \dots$$ diverges.

Is my proof sound?

• No, its not sound, because you don't explain what $k_i$s are; $s_n$ is a finite continued fraction expansion, not infinite, and I don't know how $s_n$ can converge to something that includes $n$, as $n\to\infty$ – Calvin Khor Dec 5 '19 at 4:30
• @CalvinKhor I've clarified what $k_i$s are, although I'm struggling to explain why $s_n \to \frac1{1+s_n}$... – Descartes Before the Horse Dec 5 '19 at 4:35
• $\frac{1}{n+\frac1n}<s_n<\frac1n$ – saulspatz Dec 5 '19 at 4:37

The easiest way is to say $$s_n \gt \frac 1{n+\frac 1n}\gt \frac 1{2n}$$ and the sum of $$\frac 1{2n} \gt \frac 12\log(n)$$ which diverges.
• "the sum of $\frac 1{2n} \gt \frac 12\log(n)$" Well, OK .... – zhw. Dec 5 '19 at 4:47
• I would stop at $s_n > \frac1{2n}$ – Descartes Before the Horse Dec 5 '19 at 5:14
We can reasonably assume that $$\frac{1}{2}, \frac{1}{3+\frac{1}{3}}, \cdots, \frac{1}{n+\frac{1}{n+\ldots}}$$ are all bounded above by one 1. It is fairly straightforward to prove that this is monotonic, and this is seen from the fact that $$1,\ldots,n \in \Bbb{N}.$$
We can now see that if we bound this function from below by $$\sum_{i=1}^\infty \frac{1}{i+1}$$ And it is now seen that $$\forall i\in \Bbb{N}, \frac{1}{i+1} \lt \frac{1}{i+\frac{1}{i+\ldots}}$$ because we have already said that $$\frac{1}{2}, \frac{1}{3+\frac{1}{3}}, \cdots, \frac{1}{n+\frac{1}{n+\ldots}} \lt 1$$ And so we have bounded it below by a divergent series.