0
$\begingroup$

\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.

$\endgroup$
  • $\begingroup$ Did you try wolfram alpha? $\endgroup$ – Sandeep Silwal Jan 3 at 0:49
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – HongAn Pham Jan 4 at 2:22
  • $\begingroup$ @HongAnPham. Computer Algebra System. Have a look to my edit. $\endgroup$ – Claude Leibovici Jan 4 at 4:29
  • $\begingroup$ Can you tell me what CAS did you use here? $\endgroup$ – HongAn Pham Jan 10 at 2:34
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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: $$\operatorname*{Res}_{y = -1} \frac {\Gamma(-y) \Gamma^3(1 + y)} {\Gamma^3(2 + y)} b^y = \frac {6 \ln^2 b + 12 \gamma \ln b + \pi^2 + 6 \gamma^2} {12 b}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.