Remember that little lemma from last time? The one that showed that \(q\leq p\)? Let us define a new sequence: \[p_n=\frac{a_n}{n}\] According to our lemma, \(p_n\) constitutes a decreasing sequence. Now, let us construct another sequence that bounds \(p_n\). In particular, I want the \(n^{\textrm{th}}\) term of this sequence to be less than or equal to \(p_n\). I noted that I could accomplish this by modifying \(a_n\) such that I rounded down each time, instead of merely to the nearest multiple of \(n\). Hence, I defined my new sequence as: \[b_1=p_1\] And then, each new \(b_n\) would be obtained by rounding down to the nearest \(n\) and then dividing by \(n\): \[b_{n+1}=\left\lfloor \frac{n}{n+1}b_n \right\rfloor\] I rewrote this recursive definition by splitting the floor into the difference of the actual ratio and the fractional part: \[b_{n+1}=\frac{n}{n+1}b_n-\left\{\frac{n}{n+1}b_n\right\}\] This rearranged to: \[\frac{n}{n+1}b_n-b_{n+1}=\left\{\frac{n}{n+1}b_n\right\}\] Now observe that the fractional part of any number is between \(0\) (inclusive) and \(1\) (exclusive). Hence: \[0\leq\frac{n}{n+1}b_n-b_{n+1}<1\] At this point, I observed that the first inequality in this chain could be rearranged to: \[b_{n+1}\leq\frac{n}{n+1}b_n\] So the sequence was decreasing! Next, I observed that since \(b_1>0\), then all \(b_{n+1}=\left\lfloor \frac{n}{n+1}b_n \right\rfloor\) must be at least \(0\). This implied that the sequence was bounded as well! Hence, \(\lim_{n\rightarrow\infty}{b_n}\) existed.

Since \(p_n\) is bounded below by \(b_n\) and is also decreasing, \(\lim_{n\rightarrow\infty}{p_n}\) existed as well, and all sequences of \(a_n\) must therefore converge to arithmetic sequences!

This was such a delightful problem! I do hope that my reasoning is correct. I have never quite proven something in this way before, but it makes perfect sense to me! Hooray!