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!


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.


1 Answer 1


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 $$ \bbox[lightyellow] { g(y,x) < \frac{{e^{\,x} }} {{x^{\,y - 1} }}\Gamma (y) - \frac{{\left( {y - 1} \right)}} {x}\quad \left| {\;\;1 < y,x} \right. \tag{2} }$$

  • $\begingroup$ Thanks for the reply! Perhaps I should have included the connection to Kummer's function in my description, but since the incomplete gamma function is one of its many special cases, I didn't think to include the more general function. With that said, I don't readily see how this gets me closer to my goal of closed-form bounds in terms of elementary functions (which improves on the one in my question description). Do you have any specific thoughts on the bounds? As an aside, your analysis is precisely how I moved from the original form to the form with the incomplete gamma above. $\endgroup$
    – David
    Mar 12, 2017 at 1:51
  • $\begingroup$ @David It seems that I did not get properly what you are aiming to: sorry, I restructured somewhat my answer, and if possible I will add some more about bounds in future. $\endgroup$
    – G Cab
    Mar 12, 2017 at 22:50
  • $\begingroup$ While it still isn't precisely what I'm hoping for. I appreciate the restructuring -- it is much closer to the sort of thing I'm looking for. I certainly don't want the bounty to go to waste, and this certainly has enough meat to it to have earned the rep. Thanks much! $\endgroup$
    – David
    Mar 13, 2017 at 19:14
  • $\begingroup$ @David: thanks indeed for the bounty! I did not expect it since I realized that my answer wasn't up to your needs: that means that I feel in debt to provide any further help I can find! $\endgroup$
    – G Cab
    Mar 14, 2017 at 1:15
  • $\begingroup$ No need to worry about it! As I said in the original question, I appreciate any thoughts on the topic. It is still unclear to me, to what extent what I want can be achieved. I am beginning to believe I have to further restrict my attention to particular asymptotic regimes or parameter values to get better results than what I already have. $\endgroup$
    – David
    Mar 14, 2017 at 18:36

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