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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?

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    $\begingroup$ 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 $n\to\infty$ $\endgroup$ – Calvin Khor Dec 5 '19 at 4:30
  • $\begingroup$ @CalvinKhor I've clarified what $k_i$s are, although I'm struggling to explain why $s_n \to \frac1{1+s_n}$... $\endgroup$ – Descartes Before the Horse Dec 5 '19 at 4:35
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    $\begingroup$ $\frac{1}{n+\frac1n}<s_n<\frac1n$ $\endgroup$ – saulspatz Dec 5 '19 at 4:37
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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.

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  • $\begingroup$ "the sum of $\frac 1{2n} \gt \frac 12\log(n)$" Well, OK .... $\endgroup$ – zhw. Dec 5 '19 at 4:47
  • $\begingroup$ I would stop at $s_n > \frac1{2n}$ $\endgroup$ – Descartes Before the Horse Dec 5 '19 at 5:14
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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.

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