Sum of elements of a recursive sequence Say we have $$a_n=\frac{n-1+b}{n-b}a_{n-1}$$ with $a_0>0$ and $-1<b<0$. Then it holds that $$\sum_{n=1}^{\infty}a_n=-\frac12 a_0.$$ 
I could establish this using simple manupulations with $\Gamma$ and $Beta$ funtions. I was however wondering if one could do it differently. I appreciate if you share your fun on this problem with me :-)
 A: The result depends
on two different ways
of writing the recurrence.
The first way is
$a_n
=(1-\frac{1-2b}{n-b})a_{n-1}
$.
From this
we show that
$a_m
<a_0m^{-(1-2b)}
$
or
$ma_m
<a_0m^{-2b}
$.
The second way is
$na_n-(n-1)a_{n-1}
=b(a_n+a_{n-1})
$.
Summing this,
we get
$\sum_{n=1}^{m-1}a_{n}
=\frac{(m-b)a_m}{2b}-\frac{a_0}{2}
$.
Letting $m \to \infty$,
since
$ma_m \to 0$
as $m \to \infty$
(from the first result,
since $b < 0$),
we get the desired result.
Now, the details.
$\begin{array}\\
a_n
&=\frac{n-1+b}{n-b}a_{n-1}\\
&=\frac{n-b-1+2b}{n-b}a_{n-1}\\
&=(1+\frac{2b-1}{n-b})a_{n-1}\\
&=(1-\frac{1-2b}{n-b})a_{n-1}\\
\text{so}\\
\frac{a_n}{a_{n-1}}
&=1-\frac{1-2b}{n-b}\\
\text{or}\\
\ln(a_n)-\ln(a_{n-1})
&=\ln(1-\frac{1-2b}{n-b})\\
&\lt -\frac{1-2b}{n-b}
\qquad\text{since }
\ln(1-x) < -x.\\
\text{Summing,}\\
\ln(a_m)-\ln(a_0)
&\lt -\sum_{n=1}^m\frac{1-2b}{n-b}\\
&<-(1-2b)\ln(m)\\
\text{or}\\
a_m
&<a_0m^{-(1-2b)}\\
\end{array}
$.
We have
$(n-b)a_n
=(n-1+b)a_{n-1}
$
or
$na_n-(n-1)a_{n-1}
=b(a_n+a_{n-1})
$.
Summing,
$\begin{array}\\
\sum_{n=1}^m(na_n-(n-1)a_{n-1})
&=\sum_{n=1}^m(b(a_n+a_{n-1}))\\
&=b\sum_{n=1}^ma_{n-1}+b\sum_{n=1}^ma_{n}\\
&=b\sum_{n=0}^{m-1}a_{n}+b\sum_{n=1}^ma_{n}\\
&=b(a_0+a_m+2\sum_{n=1}^{m-1}a_{n})\\
\text{or}\\
ma_m
&=b(a_0+a_m+2\sum_{n=1}^{m}a_{n})\\
\text{or}\\
\sum_{n=1}^{m-1}a_{n}
&=\frac{ma_m}{2b}-\frac{a_m}{2}-\frac{a_0}{2}\\
&=\frac{(m-b)a_m}{2b}-\frac{a_0}{2}\\
\end{array}
$
We have shown that
$ma_m
<a_0m^{2b}
\to 0
$
since $b < 0$.
Therefore,
letting $m \to \infty$,
we are done!!!!
A: I solved this using the Rising Factorial function (with partial sums using the Gamma function, although I hope to remove this by finding a simpler proof of my last step)
$$\begin{align}a_n & = \frac{n+b-1}{n-b}a_{n-1}\\
&=\frac{n+b-1}{n-b}\frac{n+b-2}{n-b-1}a_{n-2}\\
&=\frac{b+n-1}{n-b}\frac{b+n-2}{n-1-b}\cdots\frac{b}{1-b}a_0\\
&=a_0\prod_{k=1}^{n}\frac{k+b-1}{k-b}\end{align}$$
$$\implies \sum_{n=1}^{\infty}a_n = a_0\sum_{n=1}^{\infty}\prod_{k=1}^n\frac{k+b-1}{k-b}$$
$$=a_0\left(\frac{b}{1-b}+\frac{(b)(b+1)}{(1-b)(2-b)}+\frac{(b)(b+1)(b+2)}{(1-b)(2-b)(3-b)}+\cdots\right)$$
$$=\frac{b^{(1)}}{(1-b)^{(1)}}+\frac{b^{(2)}}{(1-b)^{(2)}}+\frac{b^{(3)}}{(1-b)^{(3)}}+\cdots$$
$$=a_0\left(\sum_{k=1}^{\infty} \frac{b^{(k)}}{(1-b)^{(k)}}\right)$$
$$=\color{red}{-\frac{1}{2}a_0}$$
The last part can be done by converting to a gamma representation and using partial sums, although I feel there must be an easier way... doing so is most easily done using Mathematica (as I did), but I am sure there is a more elementary proof out there. I will continue the search. I still feel that sharing most of a proof is better than nothing, so I provide this answer anyway.
