Expansion of incomplete Gamma function. I need to estimate the following quantity
$$
K(m,y) \equiv \left(\frac e m \right)^m \times \Gamma(m,(1+y)m),
$$
as $m \to \infty$, where $\Gamma(a,z)$ is the incomplete gamma function.
My question is, is there a valid expansion of the incomplete Gamma function in the range when $y \approx m^{-1/2}$ that can give us a good estimation of $K(m,y)$?

Further details:
As I need to estimate $K(m,y)$ for all $y > 0$, I looked up for uniform expansions of incomplete Gamma functions. The most promising formula that I can find is give by R. B. Paris:
$$
\Gamma(a,z)=
z^{a-1/2}e^{-z}
\left(\sqrt{\frac \pi 2}\cdot e^{\chi^2/2} \mathrm{erfc}\left(\frac{\chi}{\sqrt{2}}\right)+O(z^{-1/2})\right)
$$
where $\chi = (z-a)/\sqrt{z}$, and $\mathrm{erfc}$ is the complementary error function.
So in my case, I have
$$
K(m,y) = \left(\frac{1+y}{e^y} \right)^m \frac{1}{\sqrt{(1+y)m}}
\left(\sqrt{\frac \pi 2}\cdot e^{\chi^2/2} \mathrm{erfc}\left(\frac{\chi}{\sqrt{2}}\right)+O(m^{-1/2})\right).
$$
with $\chi = y \sqrt{m}/\sqrt{1+y}$.
When $y > m^{-1/2+\epsilon}$ or $y < m^{-1/2-\epsilon}$, i.e., when $\chi \to \infty$ or $\chi \to 0$, the above estimation is good, because we can expand $\mathrm{erfc}(\chi)$ easily.
But when $\chi$ is bounded from above and below, all I can say is that $\mathrm{erfc}(\chi) = O(1)$, which implies that
$$
K(m, y) = O(m^{-1/2}).
$$
This is not good enough for my application. I would like to get the first order approximation of $K(m,y)$ for $y$ in this range.

Even more details:
I am actually trying is to upper bound the following
$$
H(n,m) = \int_0^\infty e^{-y(n-m+1)} (1+y)^{n-m} K(m+1, y) \mathrm dy \qquad (m \to \infty),
$$
where $n > m$ and $n, m$ are both integers. 
If I use $K(m+1,y) = O(m^{-1/2})$, I will only get $H(n,m) = O(m(n-m))^{-1/2}$. But again I actually want to get the first order approximation of $H(m,n)$.
 A: From Stirling's approximation:
$$m\Gamma(m)=\Gamma(m+1)\approx\left(\frac me\right)^m\sqrt{2\pi m}$$
one can write:
$$K(m,y)=\left(\frac e m \right)^m \times \Gamma(m,(1+y)m)\approx\sqrt{\frac{2\pi}m}\times\frac{\Gamma(m,(1+y)m)}{\Gamma(m)}$$
and since $\Gamma(m,x)+\gamma(m,x)=\Gamma(m)$ then
$$K(m,y)\approx\sqrt{\frac{2\pi}m}\left(1-\frac{\gamma(m,(1+y)m)}{\Gamma(m)}\right)$$
So you only need an approximation of regularized Gamma function:
$$P(m,x):=\frac{\gamma(m,x)}{\Gamma(m)}$$
Note that $P(m,x)$ is the CDF of gamma distribution whose scale parameter is $1$. Therefore if a random variable, $x$, has a gamma distribution with shape parameter $m$, then: 
$$\mathbf{P}(x\ge m(1+y))=1-\frac{\gamma(m,(1+y)m)}{\Gamma(m)}$$
The mean and variance of a gamma distribution with unit scale are both $m$. For large $m$ the gamma distribution approaches to a normal distribution. So from a statistical point of view, you can consider a random variable $z$ with normal distribution $\mathcal N(m,m)$. Then your function becomes approximately proportional to the probability of $\mathbf P(z\ge m+my)$ or in other words:
$$z\sim\mathcal N(m,m)\rightarrow K(m,y)\approx\sqrt{\frac{2\pi}m}\mathbf P\left(\frac{z-\mu}{\sigma}\ge \sigma y\right)$$
Hence
$$K(m,y)\approx\sqrt{\frac{\pi}{2m}}\left(1-\text{erf}\left(y\sqrt{\frac m2} \right)\right)$$
A: For large $m$ with $1+\frac{k}{m}\approx 1$ and $m!\approx \left(\frac{m}{e}\right)^m\sqrt{2\pi m}$:
$\displaystyle \left(\frac{e}{m}\right)^m\Gamma(m,(1+y)m)= \left(\frac{e}{m}\right)^m (m-1)!- \left(\frac{e}{m}\right)^m \sum\limits_{k=0}^\infty\frac{(-1)^k ((1+y)m)^{k+m}}{k!(k+m)}$$\displaystyle \approx \frac{\sqrt{2\pi}}{\sqrt{m}}-\frac{(1+y)^m}{m}e^{-ym}$ 
