This has received no full solution [at StackExchange](https://math.stackexchange.com/questions/3850394/is-gammas-x-s-1-gammas-decreasing-for-real-s1-is-gammas-x-s-g).


As per <https://dlmf.nist.gov/8.10#E13> we have


$$\frac{\Gamma\left(n,n\right)}{\Gamma\left(n\right)}<\frac{1}{2}<\frac{\Gamma%
\left(n,n-1\right)}{\Gamma\left(n\right)}$$ for $n=1,2, \ldots$.


My question is: 1) show that both of these are monotone in $n$ AND 2) replacing $n$ by real $s>1$, show that these are monotone in $s$ (of course 1 follows from 2, but perhaps 1 is easier).


Note: For large $n$ both of these converge to $\frac{1}{2}$: rewriting via repeated integration by parts we get $\frac{\Gamma%
\left(n,n-1\right)}{\Gamma\left(n\right)}=\exp\{-(n-1)\}\sum\_{i=0}^{n-1} \frac{(n-1)^i}{i!}$ which is the CDF of Poisson distribution with $\lambda=n-1$ evaluated at $x=\lambda=n-1$; as $n$ increases this approaches CDF of normal with mean and variance $\lambda$, which at $x=\lambda$ is $\frac{1}{2}$. (I would be interested in alternative proofs of this fact as well.)



Is $\Gamma(s, x=s-1)/\Gamma(s)$ decreasing for real $s>1$? Is $\Gamma(s, x=s)/\Gamma(s)$ increasing?
$\newcommand\Ga\Gamma$Using integration by parts, we have
$$\Ga(n,t)=t^{n-1}e^{-t}+(n-1)\Ga(n-1,t)$$
for real $t>0$.


So, for $n\ge2$ the sign of
$$\frac{\Ga(n,t)}{\Ga(n)}-\frac{\Ga(n-1,t-1)}{\Ga(n-1)}$$ is the same as the sign of
$$t^{n-1}e^{-t}-(n-1)\int\_{t-1}^{t} x^{n-2}e^{-x}\,dx.\tag{1}$$




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Letting here $t=n-1$, we see that for $n\ge2$ the inequality $$\frac{\Ga(n,n-1)}{\Ga(n)}<\frac{\Ga(n-1,n-2)}{\Ga(n-1)}$$ can be rewritten as
$$(n-1)^{n-2}e^{-(n-1)}<\int\_{n-2}^{n-1} x^{n-2}e^{-x}\,dx,$$
which is true, because
$x^{n-2}e^{-x}$ is decreasing in $x\in(n-2,\infty)$ and hence
$x^{n-2}e^{-x}>(n-1)^{n-2}e^{-(n-1)}$ for $x\in(n-2,n-1)$. Thus, $\frac{\Ga(n,n-1)}{\Ga(n)}$ is decreasing in natural $n$ (and also in $n\in\{a,a+1,\dots\}$ for any real $a\ge1$).




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Somewhat similarly, one can show that the sign of (1) is $+$ for $t=n$, so that $\frac{\Ga(n,n)}{\Ga(n)}$ is increasing in natural $n$. Indeed, the statement that the sign of (1) is $+$ for $t=n$ can be rewritten as
$$1>\int\_{n-1}^n e^{h(x)}\,dx,\tag{2}$$
where $h(x):=(n-2) \ln x-(n-1) \ln n+n-x$. Note that $h'(x)x=n-2-x$, so that $h$ is decreasing on $[n-1,n]$. Also, $h(n-1)<1-\ln n<0$ for $n\ge3$; so, $h<0$ on $[n-1,n]$, which does yield (2) for $n\ge3$. So, $\frac{\Ga(n,n)}{\Ga(n)}$ is increasing in natural $n\ge2$. The inequality $\frac{\Ga(n,n)}{\Ga(n)}>\frac{\Ga(n-1,n-1)}{\Ga(n-1)}$ in the case $n=2$ is immediate. So, $\frac{\Ga(n,n)}{\Ga(n)}$ is increasing in all natural $n$.




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That $\frac{\Ga(n,t)}{\Ga(n)}\to1/2$ for $n\to\infty$ and $t=n+o(\sqrt n)$ follows immediately by the central limit theorem, because $\frac{\Ga(n,t)}{\Ga(n)}=P(X\_1+\cdots+X\_n>t)$, where the $X\_i$'s are iid standard exponential random variables.


