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.
$$  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc   $$
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.
$$  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  $$
