Taylor Series looks like exponential 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! =) 
 A: 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 $j = 1$.
A: 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.
