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

Question

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?

Answer 1

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.

Answer 2

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.