The sequence $a_1 = \frac{1}{2}, a_2 = 1, a_{n+1} = \frac{na_n+1}{a_{n-1}+n}$ is decreasing

Consider the sequence $\{a_n\}$ defined by $$a_1 = \frac{1}{2}, a_2 = 1, a_{n+1} = \frac{na_n+1}{a_{n-1}+n}, \forall n\ge 2.$$ Prove that $\{a_n\}_{n\ge 3}$ is decreasing.

I get the first $200$ values of $\{a_n\}$ and recognize this fact, but I cannot prove it.

Thank you very much.

From comments: it would appear that part III is not relevant for the original question. Plus, if we drop part III, the remaining induction hypotheses take effect with smaller $n.$ On the other hand, finding III is what allowed me to solve the problem; together, parts III and IV give a quite precise rate of convergence. I put in some lines of symbols to divide the parts.

Take $$ a_n = 1 + \delta_n $$

I get $$ 1 + \delta_{n+1} = \frac{n+1+ n \delta_n}{n+1+ \delta_{n-1}} $$

The induction hypotheses become extensive, and are true for $n \geq 8:$

$$ \mbox{I:} \hspace{20mm} \delta_n > 0, $$ $$ \mbox{II:} \hspace{20mm} n \delta_n < 1, $$ $$ \mbox{III:} \hspace{20mm} n \delta_n > (n-3) \delta_{n-1} > \delta_{n-1}, $$ $$ \mbox{IV:} \hspace{20mm} n \delta_n < (n-2) \delta_{n-1}. $$

We begin with part III using only the $n \delta_n > \delta_{n-1}$, $$ 1 = \frac{n+1}{n+1} < \frac{n \delta_n}{\delta_{n-1}}, $$ by the ``mediant'' inequality $$ 1 = \frac{n+1}{n+1} < \frac{n+1+ n \delta_n}{n+1+ \delta_{n-1}} <\frac{n \delta_n}{\delta_{n-1}}, $$ $$ 1 = \frac{n+1}{n+1} < 1 + \delta_{n+1} < \frac{n \delta_n}{\delta_{n-1}}, $$ and $$ 0 < \delta_{n+1}. $$ This was the induction step part I.

Context: in the setting of (simple) continued fractions, with two consecutive convergents (let's say it is for a positive irrational), when the "partial quotient" $a_n$ is equal to $1,$ the next convergent $\frac{h_n}{k_n}$ is the mediant of the previous two. $$\frac{h_n}{k_n} = \frac{a_n h_{n-1}+ h_{n-2}}{a_n k_{n-1}+ k_{n-2}} $$ Whatever $a_n$ might be, the new convergent is between the two given.

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From part II we get $$ 1 + \delta_{n+1} = \frac{n+1+ n \delta_n}{n+1+ \delta_{n-1}} < \frac{n+2}{n+1+ \delta_{n-1}} < \frac{n+2}{n+1} = 1 + \frac{1}{n+1}, $$ $$ \delta_{n+1} < \frac{1}{n+1}. $$ This was induction part II. $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Back to III, we have $$ \delta_{n-1} < \left( \frac{n}{n-3} \right) \delta_n. $$ So $$ 1 + \delta_{n+1} > \frac{n+1+ n \delta_n}{n+1+ \left( \frac{n}{n-3} \right) \delta_n}. $$ $$ n+1 + (n+1) \delta_{n+1} + \left( \frac{n}{n-3} \right) \delta_n (1 + \delta_{n+1}) > n+1 + n \delta_n. $$ $$ (n+1) \delta_{n+1} + \left( \frac{n}{n-3} \right) \delta_n (1 + \delta_{n+1}) > n \delta_n. $$ $$ 1 + \delta_{n+1} < \frac{n+2}{n+1} $$ $$ (n+1) \delta_{n+1} + \left( \frac{n}{n-3} \right) \left( \frac{n+2}{n+1} \right) \delta_n > n \delta_n. $$ For $n \geq 7,$ $$ \frac{n^2 + 2n}{n^2 -2n-3} < 2. $$ For $n \geq 7,$ $$ (n+1) \delta_{n+1} > (n-2) \delta_n $$ This was induction part III. $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

In the other direction, part IV gives $$ \delta_{n-1} > \left( \frac{n}{n-2} \right) \delta_n. $$ $$ 1 + \delta_{n+1} < \frac{n+1+ n \delta_n}{n+1+ \left( \frac{n}{n-2} \right) \delta_n}. $$ $$ n+1 + (n+1) \delta_{n+1} + \left( \frac{n}{n-2} \right) \delta_n (1 + \delta_{n+1}) < n+1 + n \delta_n. $$ $$ (n+1) \delta_{n+1} + \left( \frac{n}{n-2} \right) \delta_n (1 + \delta_{n+1}) < n \delta_n. $$ $$ (n+1) \delta_{n+1} + \left( \frac{n}{n-2} \right) \delta_n < n \delta_n. $$ $$ \frac{n}{n-2} > 1 $$ $$ (n+1) \delta_{n+1} < (n -1) \delta_n. $$ This was induction part IV. $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$