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I have a sum that comes as a special case of my original sum. I do not want to show the original sum as it is too complicated. The specific one looks like the following:

$$ G(z) = \sum_{j = 0}^{\infty} \frac{\sqrt{\pi}}{2} \frac{z^j}{j !} (1 + j)^{-\frac{3}{2}} $$

Does it belong to any known summations in literature? It should be a converging one, but I have no idea how to approach it.

Edit

In this edit, I am writing the generic sum I had in the beginning:

$$ G(z) = \frac{\Gamma(1 + n)}{\Lambda^{n+1}} \sum_{j = 0}^{\infty} \left(\frac{\frac{\Lambda}{C}}{\frac{\Lambda}{C} + j}\right)^{n+1} \frac{z^j}{j!} $$

$\Lambda, C, n > 0$

The special case above is for $\Lambda = 1, C = 1, n = 1/2$. For integer $n$, it is a hypergeometric function, but I wanted to know what it is when $n$ is not an integer.

Mathematica refuses to proceed.

Edit 2

Or, can there be a closed form of this sum that can be computed; like an asymptotic solution?

Blue
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CfourPiO
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2 Answers2

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This is not an answer; just observations.

As @TianVlašić already commented, if $$G(z) = \sum_{j = 0}^{\infty} \frac{z^j}{j !} (1 + j)^{-\frac{3}{2}}$$

$$F(z)=-\frac{\log (-z)+\Gamma (0,-z)+\gamma }{z}<G(z) <\frac{e^z-1}{z}=H(z)$$

Look at the plots of $\sqrt{F(z)\,H(z)}$ and $G(z)$ : they are extremely close to eachother.

May be, $$G(z) \sim F^a(z)\times H^{1-a}(z)$$ for some $a \in (0,1)$ $\cdots$ to be found.

Claude Leibovici
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  • Wow. Nice insight. How did you find that inequality on the first place? And, is it possible to say something similar for the general expression of $G(z)$ in my Edit ? – CfourPiO Oct 01 '23 at 08:13
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    @CfourPiO. The problem is simple with expoents $\frac 22=1$ and $\frac 42=2$. This give the two bounds. Now $\frac{1+2}3=\frac 32$. This was the idea and why I immediately tried the square root of the bounding functions.. For the other one, I am working (with little hope). Cheers :-) – Claude Leibovici Oct 01 '23 at 08:38
  • Thanks a lot. I appreciate it. In the meanwhile, I will also try the general case; to have the similar inequalities. In that case, I know that for integer exponents nearing $n+1$, the functions are generalised hyper geometric functions. $H(z)/F(z) = \frac{\Gamma(n+1)}{\Lambda^{n+1}} {(\text{round/floor}(n+1))} F{(\text{round/floor}(n+1))} (\Lambda/c_3, \Lambda/c_3, …, \Lambda/c_3; \Lambda/c_3+1, \Lambda/c_3+1, \Lambda/c_3+1,…, \Lambda/c_3+1; z)$. So, the bounding functions are; one with round and one with floor. – CfourPiO Oct 01 '23 at 18:15
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$$ S=\frac{\sqrt{\pi}}{2}\sum_{j=0}\frac{z^j}{j!}\int_0^{\infty}e^{-s(j+1)}\frac{\sqrt{s}}{\Gamma(3/2)}ds=\int_0^{\infty}e^{-s+ze^{-s}}\sqrt{s}ds=\int_0^1(-\log y)^{1/2}e^{zy}dy.$$

Edit.

$$\Gamma(\alpha)\sum_{n=0}^{\infty}\frac{1}{(c+j)^{\alpha}}\frac{z^j}{j!}=\int_0^{\infty}e^{-cs+ze^s}s^{\alpha-1}ds=\int_0^1y^{c-1}(-\log y)^{\alpha-1}e^{zy}dy.$$

Gérard Letac
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    What would you like to know about this function $z\mapsto G(z)=\int_0^1(-\log y)^{1/2}e^{zy}?$ A remarkable point is that $\log G$ is convex on $R$ (from Holder ) with its derivative valued in $(0,1).$ – Gérard Letac Oct 02 '23 at 16:05
  • Actually, I wanted to know if this function can be represented as a closed form; or a form that can be formed with special functions. The reason I wanted it because if one tries numerically to obtain a value for $G(z)$, it explodes because of the infinite sum. However, mathematically it can be proven that it is a converging function. If it can be represented as a combination of special functions like hyper-geometric, logarithmic, or Bessel, there are numerical methods to properly compute these functions. I like your answer, I appreciate it. However, I am also thinking how to generalize it. – CfourPiO Oct 03 '23 at 06:27
  • By generalize, I mean how to deal with the sum with any non-integer $n$. – CfourPiO Oct 03 '23 at 06:28
  • I have added an edit for answering to your question for general $n.$ – Gérard Letac Oct 03 '23 at 10:41
  • I appreciate the effort @Gerard . Thanks a lot for helping out really :) However, I don't see the edit, was it a technical glitch? – CfourPiO Oct 03 '23 at 12:02
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    Hum, it seems that I did a wrong click. Look : you have still a nice Laplace transform of a positive density. – Gérard Letac Oct 03 '23 at 12:35
  • Nice! How did you come up with the second line? The integral form? – CfourPiO Oct 03 '23 at 13:05
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    I suppose that you mean $\frac{1}{A^{\alpha}}=\int_0^{\infty}e^{-As}\frac{s^{\alpha-1}}{\Gamma(\alpha)}ds$ with $A=c+j.$ – Gérard Letac Oct 03 '23 at 13:44
  • Funny thing is now I realized that my original problem was the second expression you shared. I started with that actually and expanded the exponential $e^{ze^{-s}}$ into an infinite sum and then took the integral inside the sum and had the problem I shared here. Now, that I arrived at the problem again with your second expression, I realize that I could have proceeded to the third expression you shared. $\int_0^1y^{c-1}(-\log y)^{\alpha-1}e^{zy}dy.$ Is there an advantage for computing the third expression, because it is an integral from $0$ to $1$? Or, can it be reduced further? – CfourPiO Oct 03 '23 at 14:11
  • Also, in the second expression, I think it is $e^{ze^{-s}}$ instead of $e^{ze^{s}}$. Am I right? – CfourPiO Oct 03 '23 at 14:13