= Solution
For <independent and identically distributed random variables> with <distribution function> $F$, let $M_n=\max_{1\leq i\leq n}X_i$ be the <sample maximum>. In <extreme value theory>, $F$ belongs to the <maximum domain of attraction> $D(G)$ of a nondegenerate <distribution function> $G$ when there exist $a_n>0$ and $b_n\in\mathbb R$ such that
$$
\boxed{\mathbb P\left(\frac{M_n-b_n}{a_n}\leq x\right)=F(a_nx+b_n)^n\longrightarrow G(x)}
$$
at every continuity point of $G$. Thus the normalized <sample maxima> have <convergence in distribution> to $G$.
A positive measurable function $q$ has <regular variation> at infinity with index $\rho$, written $q\in\mathrm{RV}_\rho$, if
$$
\boxed{\frac{q(tu)}{q(t)}\longrightarrow u^\rho\quad(t\to\infty),\qquad u>0.}
$$
The case $\rho=0$ defines a <slowly varying function>; equivalently a <regularly varying function> is $q(t)=t^\rho\ell(t)$ with $\ell$ a <slowly varying function>.
Write $\overline F=1-F$ for the <survival function> and $x_F=\sup\{x:F(x)<1\}$ for the <right endpoint of a distribution>. For $\alpha>0$, use the following standard <extreme value theory> normalization conventions:
$$
\Phi_\alpha(x)=\begin{cases}\exp(-x^{-\alpha}),&x>0,\\0,&x\leq0,\end{cases}
\qquad
\Psi_\alpha(x)=\begin{cases}\exp(-(-x)^\alpha),&x<0,\\1,&x\geq0,\end{cases}
\qquad \Lambda(x)=\exp(-e^{-x}).
$$
They are respectively the <Fréchet distribution>, the <negative Weibull distribution>, and the <Gumbel distribution> for maxima. The <negative Weibull distribution> is supported to the left of its finite endpoint; it is not the usual positive <Weibull distribution>.
The necessary and sufficient <maximum domain of attraction> criteria are:
* For the <Fréchet distribution>,
$$
\boxed{F\in D(\Phi_\alpha)\iff x_F=\infty\ \text{and}\ \overline F\in\mathrm{RV}_{-\alpha}.}
$$
Equivalently $\overline F(tu)/\overline F(t)\to u^{-\alpha}$ for every $u>0$.
* For the <negative Weibull distribution>,
$$
\boxed{F\in D(\Psi_\alpha)\iff x_F<\infty\ \text{and}\ t\mapsto\overline F(x_F-1/t)\in\mathrm{RV}_{-\alpha}.}
$$
Equivalently $\overline F(x_F-su)/\overline F(x_F-s)\to u^\alpha$ as $s\downarrow0$, for every $u>0$.
* For the <Gumbel distribution>,
$$
\boxed{F\in D(\Lambda)\iff\exists a(t)>0:\quad
\frac{\overline F(t+a(t)x)}{\overline F(t)}\longrightarrow e^{-x}
\quad(t\uparrow x_F),\quad x\in\mathbb R.}
$$
Here $a$ is a <Gumbel auxiliary function>, and $x_F$ can be finite or infinite. For each fixed $x$, the shifted argument lies below $x_F$ eventually. This exponential tail-ratio condition, rather than <regular variation> with a fixed finite index, characterizes this case.
For the corresponding sufficient <hazard function> conditions, assume $F$ is absolutely continuous near $x_F$, its <survival function> tends to zero there, and its <probability density function> $f$ is positive there. Write $\eta(t)=f(t)/\overline F(t)$. The <Von Mises conditions for extreme values> give
$$
\boxed{\begin{array}{ll}
x_F=\infty,\quad t\eta(t)\to\alpha
&\Longrightarrow F\in D(\Phi_\alpha),\\[2pt]
x_F<\infty,\quad(x_F-t)\eta(t)\to\alpha
&\Longrightarrow F\in D(\Psi_\alpha),\\[2pt]
\displaystyle\frac{d}{dt}\left(\frac1{\eta(t)}\right)\to0
&\Longrightarrow F\in D(\Lambda).
\end{array}}
$$
The last condition additionally assumes that the reciprocal <hazard function> is continuously differentiable near the endpoint. It concerns the derivative of the reciprocal <hazard function>, not the derivative of the <hazard function> itself. Equivalently, when $f$ is differentiable, it is $\overline F(t)f'(t)/f(t)^2\to-1$.
To see the connection, taking an <integral> of the <hazard function> gives
$$
\log\frac{\overline F(v)}{\overline F(t)}=-\int_t^v\eta(u)\,du.
$$
The first two limits make the logarithmic tail ratios over multiplicative or endpoint-distance intervals tend to $-\alpha\log u$ or $\alpha\log u$, giving the required <regular variation>. For the <Gumbel distribution>, set $a(t)=1/\eta(t)$. The derivative condition makes $a(t+a(t)v)/a(t)\to1$ uniformly on bounded $v$-intervals, so
$$
\log\frac{\overline F(t+a(t)x)}{\overline F(t)}
=-\int_0^x\frac{a(t)}{a(t+a(t)v)}\,dv\longrightarrow-x.
$$
The endpoint assumptions ensure these shifts are admissible: at an infinite endpoint $a(t)/t\to0$, and at a finite endpoint $a(t)/(x_F-t)\to0$. These are sufficient smooth-tail conditions; the tail-ratio characterizations above are necessary and sufficient without differentiability assumptions. For a source using the same reciprocal-hazard convention, see https://rls.sites.oasis.unc.edu/s834-2020/ExtremeValues.pdf#page=12[Smith and Weissman's university lecture notes, equations (2.18)–(2.24)].
For the <Gamma distribution> with integer shape $m$ and rate one, let $\overline F_m(t)=\int_t^\infty u^{m-1}e^{-u}\,du/(m-1)!$. The base case is $\overline F_1(t)=e^{-t}$, and <integration by parts> gives
$$
\overline F_m(t)=\frac{t^{m-1}e^{-t}}{(m-1)!}+\overline F_{m-1}(t).
$$
Induction proves
$$
\boxed{1-F_m(t)=e^{-t}\sum_{j=0}^{m-1}\frac{t^j}{j!},\qquad t\geq0.}
$$
This <survival function> has the asymptotic form $\overline F_m(t)\sim e^{-t}t^{m-1}/(m-1)!$; for $m=1$ the relation is exact. Put $t_n=\beta_n+x$. For every fixed real $x$, $t_n\to\infty$ and $t_n/\log n\to1$. The proposed centering satisfies
$$
n e^{-\beta_n}=\frac{(m-1)!}{(\log n)^{m-1}},
$$
so
$$
n\overline F_m(\beta_n+x)
=e^{-x}\frac{(m-1)!}{(\log n)^{m-1}}
\sum_{j=0}^{m-1}\frac{(\beta_n+x)^j}{j!}
\longrightarrow e^{-x}.
$$
Only the highest-degree term contributes to the limit, with the same reasoning covering $m=1$. The <Taylor expansion> $\log(1-u)=-u+O(u^2)$ now gives
$$
n\log F_m(\beta_n+x)\longrightarrow-e^{-x},
$$
because $n\overline F_m(\beta_n+x)^2\to0$. Therefore the <Gumbel limit for gamma maxima> is
$$
\boxed{F_m(\beta_n+x)^n\longrightarrow\exp(-e^{-x}),\qquad x\in\mathbb R.}
$$
Equivalently, $M_n-\beta_n$ has <convergence in distribution> to the standard <Gumbel distribution>, with scale $a_n=1$.
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