Solving the recurrence relation $y_{n+1}=y_n+a+\frac{b}{y_n}$ I am hoping to obtain a closed-form solution or an asymptotic result for the recurrence relation
$$y_{n+1}=y_n+a+\frac{b}{y_n}$$ 
Any help would be very much appreciated!
 A: I'm going to assume that
$a, b, y_0$
are all positive.
From
$y_{n+1}
=y_n+a+\frac{b}{y_n+a}
$
it follows that
$y_{n+1}-y_n
=a+\frac{b}{y_n+a}
$
so that
$y_n
\gt na+y_0
$.
Therefore
$y_{n+1}-y_n
\lt a+\frac{b}{(n+1)a+y_0}
$.
Summing,
$y_{n}-y_0
\lt na+b\sum_{k=0}^{n-1} \frac1{(n+1)a+y_0}
\lt na+\frac{b}{a}(\ln(n)+c)
$
so
$y_n
\lt na+\frac{b}{a}\ln(n)+d
$
where $d$ is a computable constant.
From this lower bound,
we get
$\begin{array}\\
y_{n+1}-y_n
&\gt a+\frac{b}{na+\frac{b}{a}\ln(n)+d+a}\\
&= a+\frac{b}{(n+1)a+(b/a)\ln(n)+d}\\
&= a+\frac{b}{(n+1)a}\frac1{1+\frac{(b/a)\ln(n)+d}{(n+1)a}}\\
&= a+\frac{b}{(n+1)a}\frac1{1+\frac{b\ln(n)+da}{(n+1)a^2}}\\
&> a+\frac{b}{(n+1)a}(1-\frac{b\ln(n)+da}{(n+1)a^2})
\quad\text{since }\frac1{1+x} > 1-x\\
&= a+\frac{b}{(n+1)a}-\frac{b(b\ln(n)+da)}{(n+1)^2a^3}\\
\end{array}
$
Since 
$\sum \frac{\ln(n)}{n^2}$
converges,
summing this gives
$y_n
\gt na+\frac{b}{a}\ln(n) + C$
for come constant $C$.
More effort might yield
$\lim_{n \to \infty} y_n- na-\frac{b}{a}\ln(n)
$
existing
(and it probably does),
but I'll stop here.
Note:
This is a fairly standard
upper-lower bound technique
for getting asymptotics.
