The power series given by

\begin{equation} S(\alpha,n,x) = \sum_{k=0}^\infty a_k(\alpha,n)x^k,\quad n\in \pmb{N}_0,\quad x\in\pmb{R} \end{equation} where \begin{equation} a_k(\alpha,n)=\left[ \alpha k+\alpha \atop n \right] = \frac{(\alpha k+\alpha)(\alpha k+\alpha +1)\cdots (\alpha k+\alpha +n-1)}{n!},\quad \quad 0<\alpha<2\end{equation}

To find its convergence range, I tried \begin{equation} \begin{split} \lim_{k\to\infty}\frac{a_k(\alpha,n)}{a_{k-1}(\alpha,n)}&=\lim_{k\to\infty}\frac{(\alpha k+\alpha)(\alpha k+\alpha +1)\cdots (\alpha k+\alpha +n-1)}{(\alpha k)(\alpha k +1)\cdots (\alpha k +n-1)}\\ &=\lim_{k\to\infty}(1+\frac{1}{k})(1+\frac{\alpha}{\alpha k+1})\cdots(1+\frac{\alpha}{\alpha k+n-1})\\ &=1 \end{split} \end{equation}

Thus, I conclude that it converges when $|x|<1$. However, I find it's hard to test its convergence at $x=\pm 1$. I surfed the internet for hours but couldn't find a solution.

Besides, for fixed values of $\alpha$ and $n$, to find an closed form expression to the sum is even beyond my knowledge. I even don't know where to start.

I guess there may not be an explicit expression for the infinite sum, thus I tried to find approximation of it such that it can be evaluated numerically. But still I didn't succeed.

Another more challenging problem for me is to obtain the value or approximation of

\begin{equation} \sum_{n=0}^\infty [S(\alpha,n,x)]^2 \end{equation}

Can you give some reference on the above questions?

Thank you in advance!

  • $\begingroup$ Can you clarify the square-brackets notation you use? It looks to be a variation on the usual binomial coefficent, but I'm not familiar with what that's typically called. $\endgroup$ – Semiclassical Aug 22 '14 at 3:34
  • $\begingroup$ The square-brackets $\left[ n \atop k\right]$ is rising factorial. On the other hand the usual binomial coefficient is falling factorial. The notation used in the above post is the generalized rising factorial in which $n$ is allowed to be a real number. $\endgroup$ – ecook Aug 22 '14 at 6:17
  • $\begingroup$ Ah. So it's the same as multi-set coefficients, generalized to noninteger values. Note that your coefficient can be written in terms of the usual binomial coefficient as $\binom{\alpha+ k\alpha+n-1}{n}$ (which looks uglier, but may be more accessible for simplification.) $\endgroup$ – Semiclassical Aug 22 '14 at 12:40
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    $\begingroup$ You deleted math.stackexchange.com/questions/905027/… and then reposted it as a new question. The right thing to do was to edit the old version and then ask, on meta, for it to be reopened. $\endgroup$ – Gerry Myerson Aug 22 '14 at 12:46
  • $\begingroup$ Gerry Myerson Sorry for the inconvenience it made. I am new to this website and don't know the rules. Sorry $\endgroup$ – ecook Aug 22 '14 at 13:16

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