Closed-form Bound of Function related to (incomplete) Gamma function. I have a function $f:\mathbb{R}_+^3\to \mathbb{R}_+$ defined as $$ f(a,b,c) = \sum_{i=0}^{\infty}\left(\frac{a^i}{\prod_{k=0}^{i}(b+kc)}\right).$$ I am attempting to get lower bounds which are as tight as possible and preferably expressible as some closed-form combination of elementary functions. 
One such bound that I've derived, is $$f(a,b,c)\geq \left(\frac{1}{b}\right)\cdot\exp\left(\frac{a}{b+c}\right),$$
which is asymptotically tight in several interesting regimes.
A tighter bound, but one which I'm unable to put into a closed-form is $$f(a,b,c) \geq \left(\frac{1}{b}\right)\cdot \sum_{i=0}^\infty \left( \frac{a^i}{i!(b+c)^i-(i!-1)b^i}\right).$$
I've been able to show some other interesting (and potentially useful) forms of my function, namely $$f(a,b,c) = \frac{\exp\left(\frac{a}{c}\right)}{c}\cdot\left(\frac{c}{a}\right)^{b/c}\cdot \gamma\left(\frac{b}{c},\frac{a}{c}\right) = \sum_{i=0}^\infty\left(\frac{a}{c}\right)^i\cdot \left( \frac{\Gamma\left(\frac{b}{c}\right)}{c\cdot \Gamma\left(\frac{b}{c}+1+i\right) }\right),$$ where $\gamma(\cdot)$ is the lower incomplete gamma function defined by $$\gamma(x,y) = \int_0^y t^{x-1}e^{-t}dt,$$ and $\Gamma(\cdot)$ is the standard gamma function $$\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}dt.$$
I appreciate any thoughts on the matter. Thanks much!
-----UPDATE-----
One way to utilize the tighter bound is as follows:
$$\begin{align}f(a,b,c) &\geq \left(\frac{1}{b}\right)\cdot \sum_{i=0}^\infty \left( \frac{a^i}{i!(b+c)^i-(i!-1)b^i}\right)\\
&= \frac{1}{b} \left( \sum_{i=0}^\infty  \frac{\left(\frac{a}{b+c}\right)^i}{i!} + \sum_{i=0}^\infty \left(\frac{i!-1}{i!}\right)\cdot \frac{\left(\frac{a}{b+c}\right)^i}{i!(b+c)^i-(i!-1)b^i}\right) \\
&=\frac{1}{b} \left( \exp\left(\frac{a}{b+c}\right) + \sum_{i=0}^\infty \left(\frac{i!-1}{i!}\right)\cdot \frac{\left(\frac{a}{b+c}\right)^i}{i!(b+c)^i-(i!-1)b^i}\right)\\
&\geq\frac{1}{b} \left[ \exp\left(\frac{a}{b+c}\right) + \frac{n!-1}{n!} \cdot \left(\exp\left(\frac{ab}{(b+c)^2}\right)-\sum_{i=0}^{n-1}\frac{\left(\frac{ab}{(b+c)^2}\right)^i}{i!}\right)\\
+\sum_{i=0}^{n-1}  \frac{\left(\frac{i!-1}{i!}\right)\cdot\left(\frac{a}{b+c}\right)^i}{i!(b+c)^i-(i!-1)b^i}\right].
\end{align}$$
Now this inequality holds for any $n \in \mathbb{N}$ in particular for $n=2$ we have 
$$f(a,b,c) \geq \frac{1}{b} \left[ \exp\left\{\frac{a}{b+c}\right\} + \frac{1}{2} \cdot \left(\exp\left\{\frac{ab}{(b+c)^2}\right\}-1-\frac{ab}{(b+c)^2}\right)\right],$$
and since $x \geq \ln(1+x)$ this new bound is at least as tight as the old bound.
 A: Note that your function may be rewritten as
$$
\begin{gathered}
  f(a,b,c) = \sum\limits_{0\, \leqslant \,i} {\frac{{a^{\,i} }}
{{\prod\limits_{0\, \leqslant \,k\, \leqslant \,i} {\left( {b + k\,c} \right)} }}}  = \frac{1}
{c}\sum\limits_{0\, \leqslant \,i} {\frac{{\left( {a/c} \right)^{\,i} }}
{{\prod\limits_{0\, \leqslant \,k\, \leqslant \,i} {\left( {b/c + k} \right)} }}}  =  \hfill \\
   = \frac{1}
{c}\sum\limits_{0\, \leqslant \,i} {\frac{{\left( {a/c} \right)^{\,i} }}
{{b/c\prod\limits_{1\, \leqslant \,k\, \leqslant \,i} {\left( {b/c + k} \right)} }}}  = \frac{1}
{b}\sum\limits_{0\, \leqslant \,i} {\frac{{\left( {a/c} \right)^{\,i} }}
{{\prod\limits_{0\, \leqslant \,k\, \leqslant \,i - 1} {\left( {b/c + 1 + k} \right)} }}}  \hfill \\ 
\end{gathered}   
$$
so that you may concentrate in analyzing
$$
\begin{gathered}
  g(y,x) = \sum\limits_{0\, \leqslant \,j} {\frac{{x^{\,j} }}
{{\prod\limits_{0\, \leqslant \,k\, \leqslant \,j - 1} {\left( {y + k} \right)} }}}  = \sum\limits_{0\, \leqslant \,j} {\frac{{x^{\,j} }}
{{y^{\,\overline {\,j\,} } }}}  = \sum\limits_{0\, \leqslant \,j} {\frac{{\Gamma (j)}}
{{\Gamma (y + j)}}x^{\,j} }  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,j} {\frac{{j!}}
{{y^{\,\overline {\,j\,} } }}\frac{{x^{\,j} }}
{{j!}}}  = \sum\limits_{0\, \leqslant \,j} {\frac{{1^{\,\overline {\,j\,} } }}
{{y^{\,\overline {\,j\,} } }}\frac{{x^{\,j} }}
{{j!}}}  = \;_1 F_{\,1} \left( {1\,;y\,;\;x} \right) \hfill \\ 
\end{gathered} 
$$
where $y^{\,\overline {\,j\,} } $ denotes the Rising Factorial (Pochhammer symbol)
and $_1 F_{\,1} \left( {1\,;y\,;\;x} \right) $ denotes the Kummer's Confluent Hypergeometric Function.
This in turn, as you already know, is equivalent to 
$$ \bbox[lightyellow] {  
\begin{gathered}
  g(y,x) = \;_{\;1} F_{\,1} \left( {1\,;y\,;\;x} \right) = \left( {y - 1} \right)x^{\,1 - y} e^{\,x} \left( {\Gamma (y - 1) - \Gamma (y - 1,\;x)} \right) =  \hfill \\
   = \left( {y - 1} \right)x^{\,1 - y} e^{\,x} \gamma (y - 1,\;x) = \left( {y - 1} \right)x^{\,1 - y} e^{\,x} \Gamma (y - 1)\left( {1 - Q(y - 1,\;x)} \right) =  \hfill \\
   = \mathop {\lim }\limits_{p\; \to \;\infty } \;{}_2F_{\,1} \left( {1\,,p\,;y\,;\;\frac{x}
{p}} \right) \hfill \\ 
\end{gathered} 
 \tag{1} }$$
Therefore we are left to compute bounds/asymptotics for $g(y,x)$ starting either from 
the Confluent HG,  the Lower Incomplete G, the Regularized Incomplete G, or the Gaussian HG.
That premised, there is a vast literature on the subject, to explore for finding the answer that best suits to your goals.
At present I can only suggest an upper bound derived from :
$$
Q(y - 1,\;x)\;\mathop  \propto \limits_{\left| x \right|\; \to \;\infty } \;\frac{{e^{\, - x} x^{\,y - 2} }}
{{\Gamma \left( {y - 1} \right)}}\sum\limits_{0\, \leqslant \,k} {\frac{{\left( {y - 2} \right)^{\,\underline {\,k\,} } }}
{{x^{\,k} }}} 
$$
which gives
$$
\begin{gathered}
  g(y,x) = \left( {y - 1} \right)x^{\,1 - y} e^{\,x} \Gamma (y - 1)\left( {1 - Q(y - 1,\;x)} \right) =  \hfill \\
  \mathop  \propto \limits_{\left| x \right|\; \to \;\infty } \left( {y - 1} \right)x^{\,1 - y} e^{\,x} \Gamma (y - 1)\left( {1 - \frac{{e^{\, - x} x^{\,y - 2} }}
{{\Gamma \left( {y - 1} \right)}}\left( {\sum\limits_{0\, \leqslant \,k} {\frac{{\left( {y - 2} \right)^{\,\underline {\,k\,} } }}
{{x^{\,k} }}} } \right)} \right) =  \hfill \\
   = \left( {\left( {y - 1} \right)x^{\,1 - y} e^{\,x} \Gamma (y - 1) - \frac{{\left( {y - 1} \right)}}
{x}\sum\limits_{0\, \leqslant \,k} {\frac{{\left( {y - 2} \right)^{\,\underline {\,k\,} } }}
{{x^{\,k} }}} } \right) =  \hfill \\
   = \left( {\left( {y - 1} \right)x^{\,1 - y} e^{\,x} \Gamma (y - 1) - \frac{{\left( {y - 1} \right)}}
{x}\sum\limits_{0\, \leqslant \,k} {\frac{{\Gamma (y - 1)}}
{{\Gamma (y - 1 - k)x^{\,k} }}} } \right) =  \hfill \\
   = \left( {y - 1} \right)x^{\,1 - y} e^{\,x} \Gamma (y - 1)\left( {1 - \frac{1}
{{x^{\,1 - y} e^{\,x} }}\sum\limits_{1\, \leqslant \,k} {\frac{1}
{{\Gamma (y - k)x^{\,k} }}} } \right) \hfill \\ 
\end{gathered} 
$$
so that we can write, for instance
$$ \bbox[lightyellow] {  
g(y,x) < \frac{{e^{\,x} }}
{{x^{\,y - 1} }}\Gamma (y) - \frac{{\left( {y - 1} \right)}}
{x}\quad \left| {\;\;1 < y,x} \right.
 \tag{2} }$$
