Bounds on $ \sum\limits_{n=0}^{\infty }{\frac{a..\left( a+n-1 \right)}{\left( a+b \right)...\left( a+b+n-1 \right)}\frac{{{z}^{n}}}{n!}}$ I have a confluent hypergeometric function as $ _{1}{{F}_{1}}\left( a,a+b,z \right)$ where $z<0$ and $a,b>0$ and integer.
I am interested to find the bounds on the value it can take or an approximation for it. Since $$0<\frac{a..\left( a+n-1 \right)}{\left( a+b \right)...\left( a+b+n-1 \right)}<1, $$ 
I was thinking that ${{e}^{z}}$ would be an upper bound. Is that right?
I am not sure about lower bound, can this go to $-\infty$?
$$ _{1}{{F}_{1}}\left( a,a+b,z \right)=\sum\limits_{n=0}^{\infty }{\frac{a..\left( a+n-1 \right)}{\left( a+b \right)...\left( a+b+n-1 \right)}\frac{{{z}^{n}}}{n!}}\le {{e}^{z}}$$
Thanks!

The following fact helps you to find an upper bound, since $a,b>0$, then we have 
$$ \frac{(a)_n}{(a+b)_n} \leq 1,\quad \forall n\geq 0, $$
where $(a)_n$ is the Pochhammer symbol $ (a)_n = \frac{\Gamma(n+a)}{\Gamma(a)} $. The above inequality can be proved easily by using the fact
$$ (a + b)_{n} = \sum_{{j=0}}^n {n \choose j} (a)_{n-j}(b)_{j}. $$  
