Taylor Series looks like exponential

Question

guys!

I need a little help.

I have this series

$$ \sum_{k \ge 0} \frac{\Gamma(j)}{\Gamma(j+k/2)}(-t)^k $$

where $j \in \mathbb{N}$.

I need to know if the limit of this function when $t$ goes to infinity goes to zero.

If you use the definition of the gamma function you get

$$ \sum_{k \ge 0} \frac{j!}{(j+k/2)!}(-t)^k \le \sum_{k \ge 0} \frac{(-t)^k}{(\frac{k}{2})!}$$

the series in the right side looks like the Taylor series for some kind of exponential. At first sight my guess was $exp(-t^2)$ but I'm wrong.

Any ideas? Maybe I don't need to know the function, but is it possible to know if this series goes to zero when $t$ goes to infinity?

Thanks! =)

Answer 1

It can be expressed in terms of confluent hypergeometric function of the first kind:

$$M(a,b,z) = {}_1\!F_1(a;b;z) = \sum_{k=0}\frac{(a)_k}{(b)_k}\frac{z^k}{k!}$$ We have $$\begin{align} \sum_{k=0}^\infty \frac{\Gamma(j)}{\Gamma(j+\frac{k}{2})} (-t)^k & = \sum_{k=0}^\infty \left( \frac{\Gamma(j)}{\Gamma(j+k)} - \frac{\Gamma(j)t}{\Gamma(j+ \frac12+k)}\right) (t^2)^k\\ & = \sum_{k=0}^\infty \left( \frac{(t^2)^k}{(j)_k} - \frac{\Gamma(j)t}{\Gamma(j+\frac12)}\frac{(t^2)^k}{(j+\frac12)_k} \right)\\ &= M(1,j,t^2)-\frac{\Gamma(j)t}{\Gamma(j+\frac12)} M\left(1,j+\frac12,t^2\right) \end{align}\tag{*1} $$

For large $|z|$, the asymptotic behavior of $M(a,b,z)$ is given by

$$M(a,b,z) \asymp \Gamma(b)\left(\frac{e^z z^{a-b}}{\Gamma(a)} + \frac{(-z)^{-a}}{\Gamma(b-a)}\right) $$ where the powers of $z$ are taken using $-\frac32\pi < \arg z \le \frac12\pi$.

Substitute this in $(*1)$, you should get something like

$$(*1) \asymp \frac{\Gamma(j)}{t^2}\left(\frac{t}{\Gamma(j-\frac12)}-\frac{1}{\Gamma(j-1)}\right)\tag{*2}$$

Update

To see why the $e^{t^2}$ terms cancel out, let us first consider the case $j > 1$. When $j > 1$, we can use following integral representation of $M(a,b,z)$ which is valid when $\Re b > \Re a > 0$:

$$M(a,b,z) = \frac{\Gamma(b)}{\Gamma(a)\Gamma(b-a)}\int_0^1 e^{zu} u^{a-1}(1-u)^{b-a-1} du$$

We find $$\begin{align} (*1) &= \frac{\Gamma(j)}{\Gamma(j-1)}\int_0^1 e^{t^2u}(1-u)^{j-2} du - \frac{\Gamma(j)t}{\Gamma(j-\frac12)}\int_0^1 e^{t^2u} (1-u)^{j-3/2} du\\ &= \Gamma(j)e^{t^2}\int_0^1 e^{-t^2u}\left(\frac{u^{j-2}}{\Gamma(j-1)} - \frac{tu^{j-3/2}}{\Gamma(j-\frac12)}\right) du\\ &= \frac{\Gamma(j)e^{t^2}}{t^{2(j-1)}}\int_0^{t^2} e^{-u}\left(\frac{u^{j-2}}{\Gamma(j-1)} - \frac{u^{j-3/2}}{\Gamma(j-\frac12)}\right) du \end{align} $$ Here comes to the key point. Using the integral representation of gamma functions, we know the integral in last expression goes to $0$ as $t^2 \to \infty$. As a result, we get

$$\begin{align} (*1) &= \frac{\Gamma(j)e^{t^2}}{t^{2(j-1)}}\int_{t^2}^\infty e^{-u}\left( \frac{u^{j-3/2}}{\Gamma(j-\frac12)} - \frac{u^{j-2}}{\Gamma(j-1)} \right) du\\ &= \frac{\Gamma(j)}{t^2}\int_0^\infty e^{-u}\left[ \frac{t\left(1+\frac{u}{t^2}\right)^{j-3/2}}{\Gamma(j-\frac12)} - \frac{\left(1 + \frac{u}{t^2}\right)^{j-2}}{\Gamma(j-1)} \right] du \end{align}\tag{*3}$$ As one can see, the $e^{t^2}$ term from the even and odd part of $(*1)$ cancel each other!

Notice for fixed $s$ and large $t$, we have $$\int_0^\infty e^{-u} \left(1 + \frac{u}{t^2}\right)^s du \asymp 1 + \frac{s}{t^2} + \cdots$$ This means the leading asymptotic behavior of $(*1)$ is indeed given by $(*2)$. Since the integral in RHS of $(*3)$ is well behaved as $j \to 1$, $(*2)$ also works for the case $j = 1$.

Answer 2

Expanding on what David H commented, split the series into odd and even terms and use the duplication formula for Gamma on the odd terms. You get something of the form \begin{equation*} \sum_{l \geq 0} \frac{\Gamma(j)}{\Gamma(j+l)}t^{2l} - 2t\pi^{\frac{1}{2}} \sum_{l \geq 0} 2^{-2(j+l)} \frac{\Gamma(j)\Gamma(j+l)}{\Gamma(2j+2l)} t^{2l}. \end{equation*} The first series can easily be bounded from below (in the tail) as \begin{equation*} \sum_{l > j} \frac{t^{2l}}{(j+l)\cdots(j+1)} \geq \sum_{l > j}\frac{t^{2l}}{2^ll!} = \exp(\frac{t^2}{2}) + O_j(1). \end{equation*} For the second series, rewriting as factorials (so the product of gamma functions gives $\frac{j!}{(j+l)!}\binom{2j+2l}{j+l}^{-1}$), you get an upper bound (very crudely) as \begin{equation*} 2t\pi^{\frac{1}{2}} \sum_{l \geq 0} 2^{-2(j+l)} \frac{\Gamma(j)\Gamma(j+l)}{\Gamma(2j+2l)} t^{2l} \leq 2^{1-2j}t\pi^{\frac{1}{2}} \sum_{l \geq 0} (t/4)^{2l}\frac{j!}{(j+l)!} \leq 2^{1-2j}t\pi^{\frac{1}{2}} \sum_{l \geq 0} \frac{(t/4)^{2l}}{l!} \ll_j t\exp(t^2/16) \leq \exp(t^2/8) \end{equation*} so the difference is definitely bounded below by some function going off to infinity.