# Convergent or divergent series? Asymptotic of the series?

\begin{align} A&=\sum_{n\ge 0} \frac{(-1)^n}{n!} \frac{b^n}{(n+1)^3}\\ &= \sum_{n=0}^{\infty}\frac{(-1)^{n}b^{n}}{n!}\frac{1}{(1+n)^{3}} = 1\,{}_{3}F_{3}(1,1,1;2,2,2;-b) \end{align} The series is convergent or divergent when b ~ 10^6? Can we survey the asymptotic of this series when b large? I have use the wolfarm alpha and see that it come to zero for b large! However it's not help for my problem. I need an approximation when b large for this series so any suggestions for me to do that? $${}_{3}F_{3}(1,1,1;2,2,2;-b)$$ is the generalized hypergeometric function.

• Did you try wolfram alpha? – Sandeep Silwal Jan 3 at 0:49

I do not see where the factor $$2$$ is coming from. To me $$A=\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \frac{b^n}{(n+1)^3}=\, _3F_3(1,1,1;2,2,2;-b)$$

Playing with a CAS (Computer Algebra System) I arrived to $$A=\frac{6 \log ^2(b)+12 \gamma \log (b)+(\pi ^2+6 \gamma ^2)}{12 b}-\frac{e^{-b}}{b^3}\left(1-\frac{3}{b}+\frac{11}{b^2}-\frac{50}{b^3}+\frac{274}{b^4}+\cdots \right)$$ and then the asymptotics $$A\simeq\frac{6 \log ^2(b)+12 \gamma \log (b)+(\pi ^2+6 \gamma ^2)}{12 b}$$ which does not seem bad at all (the table was generated for twenty significant figures). $$\left( \begin{array}{ccc} b & \text{approximation} & \text{exact}\\ 10 & 0.49690932360120875139 & 0.49690928832388243555 \\ 20 & 0.36027227594404510777 & 0.36027227594382029864 \\ 30 & 0.29121174059860656362 & 0.29121174059860656046 \\ 40 & 0.24805627039169795554 & 0.24805627039169795554 \\ 50 & 0.21798197906031610042 & 0.21798197906031610042 \\ 60 & 0.19557007518743034227 & 0.19557007518743034227 \\ 70 & 0.17808871303080732581 & 0.17808871303080732581 \\ 80 & 0.16399386407966419640 & 0.16399386407966419640 \\ 90 & 0.15233955734129875868 & 0.15233955734129875868 \\ 100 & 0.14251028587174600483 & 0.14251028587174600483 \\ 200 & 0.090417056176257986847 & 0.090417056176257986847 \\ 300 & 0.068493119493293235309 & 0.068493119493293235309 \\ 400 & 0.055990617243116989201 & 0.055990617243116989201 \\ 500 & 0.047773804098897067512 & 0.047773804098897067512 \\ 600 & 0.041903030844139627443 & 0.041903030844139627443 \\ 700 & 0.037469669949770873800 & 0.037469669949770873800 \\ 800 & 0.033986919466450244270 & 0.033986919466450244270 \\ 900 & 0.031168657855126580756 & 0.031168657855126580756 \\ 1000 & 0.028834862048815511463 & 0.028834862048815511463 \\ 2000 & 0.017131637531050345297 & 0.017131637531050345297 \\ 3000 & 0.012553805862832047429 & 0.012553805862832047429 \\ 4000 & 0.010043035272326017788 & 0.010043035272326017788 \\ 5000 & 0.0084353206512719321002 & 0.0084353206512719321002 \\ 6000 & 0.0073085551020333550676 & 0.0073085551020333550676 \\ 7000 & 0.0064704608754287297893 & 0.0064704608754287297893 \\ 8000 & 0.0058201825163032066414 & 0.0058201825163032066414 \\ 9000 & 0.0052994357876654838989 & 0.0052994357876654838989 \\ 10000 & 0.0048720593620934820453 & 0.0048720593620934820453 \end{array} \right)$$

Edit

If you expand the first terms (say, up to $$7$$) $$\sum_{n=0}^{7} \frac{(-1)^n}{n!} \frac{b^n}{(n+1)^3}=1-\frac{b}{8}+\frac{b^2}{54}-\frac{b^3}{384}+\frac{b^4}{3000}-\frac{b^5}{25920}+\frac{b^6}{246960}-\frac{b^7}{2580480}$$ you should find that sequence $$\{1, 8, 54, 384, 3000, 25920, 246960, 2580480, 29393280\}$$ is $$A002775$$ in $$OEIS$$; please read the very first comment in the linked page.

• sorry but I don't understand what CAS is? and How can you limit the expansion like that, can you do it step by step please? – HongAn Pham Jan 4 at 2:22
• @HongAnPham. Computer Algebra System. Have a look to my edit. – Claude Leibovici Jan 4 at 4:29
• Can you tell me what CAS did you use here? – HongAn Pham Jan 10 at 2:34

That series is in fact convergent for all $$b\in\mathbb{R}$$; you are dividing by a factorial, and generally speaking all factorials dominate all exponentials. You can learn more about this phenomenon by reading up on Stirling's approximation.

The parameters of the hypergeometric function come from $$(1)_n/(2)_n = 1/(n + 1)$$. Here is a sketch of how an asymptotic estimate can be derived. We have $${_3\hspace{-1px}F_3}(1, 1, 1; 2, 2, 2; -b) = G_{3, 4}^{1, 3} \left( b \middle| {0, 0, 0 \atop 0, -1, -1, -1} \right) = \frac 1 {2 \pi i} \int_C \frac {\Gamma(-y) \Gamma^3(1 + y)} {\Gamma^3(2 + y)} b^y dy.$$ The poles of $$\Gamma(-y)$$ are on the rhs of the contour $$C$$. We can extend the contour indefinitely to the left to make $$y$$ large everywhere on $$C$$, replace $$\Gamma(-y)$$ with its asymptotic approximation and estimate the integral by applying the steepest descent method. The saddle point will be at $$y = -b$$, thus the integral will be exponentially small in $$b$$. Now we just need to add the contribution from the pole at $$y = -1$$, which appeared when moving the contour: $$\operatorname*{Res}_{y = -1} \frac {\Gamma(-y) \Gamma^3(1 + y)} {\Gamma^3(2 + y)} b^y = \frac {6 \ln^2 b + 12 \gamma \ln b + \pi^2 + 6 \gamma^2} {12 b}.$$