Necessary and sufficient condition for weak convergence of gamma distribution Let $(X_n)_n$ be a sequence of random variable such that $f_{X_n}(x)=\frac{1}{\Gamma(\alpha_n)}\lambda_n^{\alpha_n}x^{\alpha_n-1}e^{-\lambda_nx}1_{]0,+\infty[}(x),$ where $\alpha_n>0,\lambda_n>0.$

*

*Suppose that $\alpha_n=1,\forall n \in \mathbb{N}.$ Find a necessary and sufficient condition on $(\lambda_n)_n$ such that $(X_n)_n$ converges in distribution.


*More generally, find a necessary and sufficient condition on $\alpha_n,\lambda_n$ so that $(X_n)_n$ converges in distribution.
The first part is easy, it converges in distribution if and only if $0<\liminf_n\lambda_n=\limsup_n\lambda_n$.
Concerning part 2), is it true that a condition of weak convergence is the convergence of $(\alpha_n)$ and $(\lambda_n)$?
 A: "Is it true that a condition of weak convergence is the convergence of $\alpha_n$ and $\lambda_n$?" The answer is negative.
Counterexample:
Suppose $\lambda_n \to +\infty$, $\alpha_n = \bar{o}(\lambda_n)$, $n \to \infty$, and let $X_n$ have distribution $\Gamma(\lambda_n, \alpha_n)$. We have
$$\mathbb{E} e^{i t X_n}  = \Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{-\alpha_n} = e^{-\frac{\alpha_n}{\lambda_n} \cdot \lambda_n \ln\bigl( 1 - \frac{it}{\lambda_n} \bigr)}.$$
Hence,
$$\lim_{n \to \infty} \mathbb{E} e^{i t X_n} =
\lim_{n \to \infty} e^{-\bar{o}(1) (-it)(1+\bar{o}(1))} = 1 = \mathbb{E} e^{i t \cdot 0},
$$
that is $X_n \to 0$ in distribution.
So we may take, for example, $\lambda_n = n$ and $\alpha_n = 2 + (-1)^n$ for our counterexample.
If we suppose that $\alpha_n$ and $\lambda_n$ are separated from zero and $\infty$, that is $0 < c \le \alpha_n, \lambda_n \le C < \infty$, than convergence of $X_n$ is equaivalent to pointwise convergence of functions  $\Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{-\alpha_n} $ for each $t \in \mathbb{R}$. In this case we may show that "there is convergence in distribution" iff "$\alpha_n$ and $\lambda_n$ are convergent" (and in this case $lim_n X_n \sim \Gamma(\lim_n \lambda_n, \lim_n \alpha_n)$).  The "if-part" follows immediately from the convergence of characteristic functions.
To prove the second part, let us take a convergent subsequence $\alpha_{n_k}$. We know that $X_n$ has a limit, hence $\exists \lim_n \Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{\alpha_n}$. Put $t=1$. Sequences $\Bigl(1 - \frac{i}{\lambda_{n_k}} \Bigr)^{\alpha_{n_k}}$ and $\alpha_{n_k}$ are convergent. Hence, for some finite $\lambda, \alpha>0$ we have $\lambda_{n_k} \to \lambda$, $\alpha_{n_k} \to \alpha$  and hence $$\lim_n \Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{\alpha_n} \to \Bigl(1 - \frac{it}{\lambda} \Bigr)^{\alpha}.$$
If there is another convergent subsequence $\alpha_{n_k^*} \to \alpha^*$, we have (as above): $\lambda_{n_k^*} \to \lambda^*$ and $$\lim_n \Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{\alpha_n} \to \Bigl(1 - \frac{it}{\lambda^*} \Bigr)^{\alpha^*}.$$ Moreover, $\alpha^*, \lambda^* > 0$. We need to prove that $\alpha = \alpha^*$ and  $\lambda = \lambda^*$.
It's easy to see that
$$\Bigl(1 - \frac{it}{\lambda} \Bigr)^{\alpha} = \Bigl(1 - \frac{it}{\lambda^*} \Bigr)^{\alpha^*}$$
for all $t \in \mathbb{R}$. Put $C = \frac{\alpha}{\alpha^*} > 0$. We have $\Bigl(1 - \frac{it}{\lambda} \Bigr)^{C} = 1 - \frac{it}{\lambda^*} $. Put $u(t) = 1 - \frac{it}{\lambda} $. We have: $u^C = $ linear function of $u$, $u \in \mathbb{R}$. Hence, $C=1$ and $\alpha = \alpha^*$.
Moreover, $\Bigl(1 - \frac{it}{\lambda} \Bigr)^{C} = 1 - \frac{it}{\lambda^*} $, so $\lambda = \lambda^*$. The proof is finished.
For case when $\alpha_n$ and $\lambda_n$ are not separated from $0$ both we have the following theorem.
Theorem 1.
If $\alpha_n \ge c > 0$ and $\lambda_n \to 0$ then $X_n$ diverges.
If $\alpha_n \to 0$ and $\lambda_n \ge c > 0$ then $X_n \overset{w}{\to} 0$.
If $\alpha_n \to 0$ and $\lambda_n \to 0$ then
$$X_n \overset{w}{\to} \Longleftrightarrow X_n \overset{w}{\to} 0 \Longleftrightarrow \lim_{n \to \infty} \alpha_n \ln \lambda_n = 0.$$
Let us prove the theorem. If $\xi \sim \Gamma( \lambda, \alpha)$ then
\begin{gather}
\Bigl( 1 - \frac{it}{\lambda} \Bigr)^{-\alpha} = e^{-\alpha \ln(1 - \frac{it}{\lambda} )} = e^{-\alpha  \Bigl( \ln \sqrt{1^2 + \frac{t^2}{\lambda^2}} + i \cdot arg(-\frac{t}{\lambda}) \Bigr) } = \nonumber \\
= e^{ - \frac{\alpha}2 \ln \bigl(1 + \frac{t^2}{\lambda^2} \bigr)} \cdot e^{i \alpha arctg \frac{t}{\lambda} } = {\Bigl( 1 + \frac{t^2}{\lambda^2}\Bigr)}^{-\frac{\alpha}2} \cdot e^{i \alpha arctg \frac{t}{\lambda} }.\nonumber
\end{gather}
Hence $$Ee^{itX_n} = {\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)}^{-\frac{\alpha_n}2} \cdot e^{i \alpha_n arctg \frac{t}{\lambda_n} }.$$
From Levy’s continuity theorem we have
$$ \Bigl( X_n \overset{w}{\to} \Bigr)  \Longleftrightarrow \Bigl(  \forall t \in \mathbb{R} \ \ \exists \lim_n Ee^{itX_n} = \phi(t) \text{and  } \phi(t) \text{  is continious at } 0 \Bigr).$$
If $\alpha_n \ge c > 0$ and $\lambda_n \to 0$ then $\lim_n |Ee^{itX_n}| = {\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)}^{-\frac{\alpha_n}2} \to I_{t=0}$, where $I$ is an indicator function. If $X_n \overset{w}{\to} X$ then $\lim_n |Ee^{itX_n}| = |Ee^{itX}|$ is continuous (as it is the module of the characteristic function). But $I_{t=0}$ is not continuous.  Hence $X_n$ diverges. The first statement of the theorem is proved.
Here and everywhere below we will assume that $\alpha_n \to 0$. Thus $\alpha_n arctg \frac{t}{\lambda_n}  = \text{ bounded } \cdot \alpha_n = \bar{o}(1)$, $n \to \infty$. Therefore $e^{i \alpha_n arctg \frac{t}{\lambda_n} } = 1 + \bar{o}(1).$
Put $\phi_n(t) ={\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)}^{-\frac{\alpha_n}2}$. We proved that
\begin{gather} \Bigl( X_n \overset{w}{\to} \Bigr)  \Longleftrightarrow \Bigl(  \forall t \in \mathbb{R} \exists \lim_n \phi_n(t) = \phi(t) \text{ and } \phi(t) \text{ is continious at  } 0 \Bigr).          (1)
\end{gather}
If $\alpha_n \to 0$ and $\lambda_n \ge c > 0$ then $\phi_n(t) \to  1$ and moreover $Ee^{itX_n} \to 1$, hence $X_n \overset{w}{\to} 0$.  The second statement of the theorem is proved.
Here and everywhere below we will assume that $\lambda_n \to 0$. For $t \ne 0$ we have
\begin{gather}
\phi_n(t) =  e^{-\frac{\alpha_n}2 \ln {\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)} } =  e^{-\frac{\alpha_n}2 (1 + \bar{o}(1))\ln {\Bigl(\frac{t^2}{\lambda_n^2}\Bigr)} } = \nonumber \\
= e^{-\alpha_n (1 + \bar{o}(1)) ( \ln t - \ln \lambda_n  )} = e^{\alpha_n \ln \lambda_n  (1 + \bar{o}(1))}
\end{gather}
Hence for $t \ne 0$  we have
$$ \exists lim_n \phi_n(t) \Longleftrightarrow  \exists lim_n \alpha_n \ln \lambda_n = A \in [-\infty, 0]$$
and if this condition holds true then $\phi(t) =  lim_n \phi_n(t) = e^{A}$, $t \ne 0$. Obviously $lim_n \phi_n(0) = lim_n 1 = 1$.
So
\begin{gather}
\Bigl(  \forall t \in \mathbb{R} \ \ \exists \lim_n \phi_n(t) = \phi(t) \text{ and } \phi(t) \text{ is continious at } 0 \Bigr) \Longleftrightarrow \nonumber \\
\exists lim_n (\alpha_n \ln \lambda_n) = A \text{ and } A=0.        (2)
\end{gather}
From (1) and (2) we get that $X_n$ converges iff $ \lim_{n \to \infty} \lambda_n \ln \alpha_n = 0$. Moreover, if $ \lim_{n \to \infty} \lambda_n \ln \alpha_n = 0$ then $Ee^{itX_n} =  \phi_n(t) \cdot e^{i \alpha_n arctg \frac{t}{\lambda_n} } \to 1 = Ee^{it\cdot 0}$, hence $X_n \overset{w}{\to} 0$.
Theorem is proved.
A: This comment is the second part of my previous comment,
I write it separately for the convenience of typing.
What have we proved already? In case, when $\alpha_n \to 0$ or $\lambda_n \to 0$ or they both tend to zero, the answer is given by the previous theorem. Suppose now that $\alpha_n$ and  $\lambda_n$ are separated from zero. A case when both of them are separated from $\infty$  was considered also --- in this case we proved that $ \Bigl( X_n \overset{w}{\to} \Bigr) \Longleftrightarrow \exists \lim_n \lambda_n$ and  $\exists \lim_n \alpha_n$. Now we will deal with the last case, when $\alpha_n$ and  $\lambda_n$ are separated from zero and at least one of them is not separated from $\infty$.
Theorem 2.
Suppose that $\alpha_n\ge c_1 > 0$ and $\lambda_n \ge c_1 > 0$.
If $\alpha_n \le c_2$ and $\lambda_n \to \infty$ then $X_n  \overset{w}{\to} 0$.
If $\alpha_n \to \infty$ and $\lambda_n  \le c_2$ then $X_n$ diverges.
If  $\alpha_n \to \infty$ and $\lambda_n \to \infty$ then
$$X_n  \overset{w}{\to} \Longleftrightarrow \exists \lim_n \frac{\alpha_n}{\lambda_n},$$
and if $X_n$ is convergent then $X_n  \overset{w}{\to} \lim_n \frac{\alpha_n}{\lambda_n}$.
Let us prove the theorem 2.
The first and the second statements may be proved similarly to the proof of Theorem 1.
Consider the most interesting case: $\alpha_n \to \infty$ and $\lambda_n \to \infty$. Here and everywhere below we will assume that this condition holds.
As above, we have
$$Ee^{itX_n} = {\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)}^{-\frac{\alpha_n}2} \cdot e^{i \alpha_n arctg \frac{t}{\lambda_n} } = e^{- \frac{\alpha_n}2 \cdot \frac{t^2}{\lambda_n^2}(1+\bar{o}(1))}  \cdot e^{i  t \frac{\alpha_n}{\lambda_n} (1+\bar{o}(1))}. $$
Suppose that $X_n$ is convergent. Hence $|Ee^{itX_n}|$ converges to a continuous function. But $|Ee^{itX_n}| = e^{- \frac{\alpha_n}2 \cdot \frac{t^2}{\lambda_n^2}(1+\bar{o}(1))}  $ thus $ \frac{\alpha_n}{\lambda_n^2} \to A \in [0, \infty)$. Moreover, $Ee^{itX_n}$ converges to a continuous function. Hence $e^{i  t \frac{\alpha_n}{\lambda_n} (1+\bar{o}(1))}$ converges to a continuous function. It means that $\xi_n = \frac{\alpha_n}{\lambda_n}$ converges weakly to some $\xi$. Put $F_{\xi_n}(x) = P(\xi_n \le x)$, $F_{\xi}(x) = P(\xi_n \le x)$. We know that $F_{\xi_n}(x) \to F_{\xi}(x) $ for all $x$ except at most countable set. Hence, $F_{\xi}(x)$ is a distribution function of some constant. It's easy to see that $\exists \lim_n \frac{\alpha_n}{\lambda_n} < \infty$ and this limit is equal to $\xi$. So we proved that
$$ \Bigl( X_n  \overset{w}{\to} \Bigr) \Longrightarrow \Bigl(  \exists \lim_n \frac{\alpha_n}{\lambda_n} \Bigr).   $$
Now suppose that $\exists \lim_n \frac{\alpha_n}{\lambda_n} = B < \infty$. Thus $\exists \lim_n \frac{\alpha_n}{\lambda_n^2} = 0$.
We know that $Ee^{itX_n} = e^{- \frac{\alpha_n}2 \cdot \frac{t^2}{\lambda_n^2}(1+\bar{o}(1))}  \cdot e^{i  t \frac{\alpha_n}{\lambda_n} (1+\bar{o}(1))}$ hence $Ee^{itX_n} = e^{i t B (1+\bar{o}(1))}$.
Thus
$$\Bigl(\exists \lim_n \frac{\alpha_n}{\lambda_n} = B \Bigr)  \Longrightarrow \Bigl(X_n  \overset{w}{\to} B\Bigr),$$
q.e.d.
Finally, let us present an informal proof of the biggest part of the theorem 2: the case when $\alpha_n \to \infty$.
It's well known that $\Gamma(\lambda, m)$ is a sum of $m$ independent exponential random variables $\exp{\lambda}$ with parameter $\lambda$. The following symbolic equality is fulfilled: $\exp{\lambda} = \frac{\exp{1}}{\lambda}$, it means that $\lambda$ is a scale parameter. Hence in case  $\alpha_n \to \infty$. we have
\begin{gather}
X_n  \overset{d}{=}  \Gamma(\lambda_n, \alpha_n) \approx \sum_{i=1}^{[\alpha_n]} \exp{\lambda_n} = \sum_{i=1}^{[\alpha_n]} \frac{\exp{1}}{\lambda_n} \nonumber =\\
=\frac{\sum_{i=1}^{[\alpha_n]} \exp{1} }{[\alpha_n]} \cdot \frac{[\alpha_n]}{\lambda_n} = \text{S.L.L.N.} = (1+\bar{o}(1)) \cdot \frac{[\alpha_n]}{\lambda_n} \approx \frac{\alpha_n}{\lambda_n}. \nonumber
\end{gather}
Now it's easy to see that if $\lambda_n \le c_2$ then $X_n$ diverges and if $\lambda_n \to \infty$ then $\Bigl(X_n  \overset{w}{\to}\Bigr)$ iff  $\exists \lim_n \frac{\alpha_n}{\lambda_n}$. Moreover, if $X_n$ is convergent then $X_n  \overset{w}{\to} \lim_n \frac{\alpha_n}{\lambda_n}$.
Now it's easy to see construct "bad" sequences $\alpha_n$ and $\lambda_n$ such that $X_n$ converges. For example, put
\begin{gather}
\alpha_{3n} = \frac1{n}, \lambda_{3n} = 2+(-1)^n,\\
\alpha_{3n+1} = 2+(-1)^n, \lambda_{3n+1} = n,\\
\alpha_{3n+2} = \frac1{n}, \lambda_{3n+2} = \frac1{n}.
\end{gather}
It follows from Theorem 1 and Theorem 2 that $X_n \overset{d}{\to} 0$.
A: Let $\{\mu_n\overset{\text{d}}{=}\Gamma(\alpha_n,\lambda_n), n\ge 1\}$ be a sequence of Gamma-distributions.
The necessary and sufficient conditions for weak convergence of Gamma-distribution
could be expressed as follows:

*

*The necessary and sufficient conditions of $ \mu_n\Rightarrow\delta_0 $ (where $\delta_0$ is a single point probability distribution concentrated at $0$) are
\begin{equation*}
   \lim_{n\to\infty}\alpha_n\log\Big(1+\frac1{\lambda_n}\Big)= 0. \tag{1}
\end{equation*}


*The necessary and sufficient conditions of $ \mu_n\Rightarrow\delta_c$ (where $\delta_c$ is a single point probability distribution concentrated at $c>0$) are
\begin{equation*}
 \lim_{n\to\infty}\lambda_n=\infty, \qquad  
 \lim_{n\to\infty}\frac{\alpha_n}{\lambda_n}= c.  \tag{2}
\end{equation*}


*The necessary and sufficient conditions of $ \mu_n\Rightarrow\mu\ne\delta_c(c\ge 0)$ are
\begin{equation*}
 \lim_{n\to\infty}\alpha_n=\alpha\in(0,\infty),\qquad
\lim_{n\to\infty}\lambda_n=\lambda\in(0,\infty).  \tag{3}
\end{equation*}
If $\mu$ is a probibilty distribution on $\mathbb{R}_+=[0,\infty)$, then define the Laplace
transform of $ \mu $ as follows:
$$ L(u)=\int_0^\infty e^{-ux}\mu(dx),\qquad u>0,  
$$
and
$$ \psi(u) =-\log L(u).
$$
(For the Laplace transform of distribuions please refer to the
W. Feller, An introduction to Probability Theory and Its Applications(2nd Ed.),
Vol.II John Wiley & Sons, Inc.(1971). Ch.13.)
If $\mu=\Gamma(\alpha,\lambda)$ is a $\Gamma$-distribution, then it is a infinitively divisible distribution on $\mathbb{R}_+$, and its  Laplace transform is
$$ L(u)=\Big(1+\frac{u}{\lambda}\Big)^{-\alpha}.
$$
and
\begin{gather*}
 \psi(u)=-\log L(u)=\alpha\log\Big(1+\frac{u}{\lambda}\Big). \tag{4}\\
 \alpha=\frac{\psi(u)}{\log\Big(1+\dfrac{u}{\lambda}\Big)}\qquad
 \lambda=\frac{u}{e^{\psi(u)/\alpha}-1}.  \tag{5}
\end{gather*}
In particular,
$$ \psi_{\delta_0}(u)=0,\qquad \psi_{\delta_c}(u)=cu.
$$
For $\mu_n\overset{w}{\to}\mu^\ast$, iff the following limit is true(c.f. Feller's book),
\begin{gather*}
 L_n(u)\to L^\ast(u),\qquad \forall u\ge0,\\
 \begin{aligned}
  \psi_n(u)&=-\log L_n(u)=\alpha_n\log\Big(1+\frac{u}{\lambda_n}\Big)\\
  &\to\psi^\ast(u)=\log L^\ast(u),\qquad 
  \forall u\ge0.
 \end{aligned} \tag{6}
\end{gather*}
Using the similar method for characteristic functions we could also deduce the convergence in (6) is uniform at $ u=0 $, that is
\begin{equation*}
 \lim_{n\to\infty}\psi_n(u_n)=0.\qquad \forall u_n\to0. \tag{7}
\end{equation*}
Firstly, to prove the sufficiency of (1), it is suffice to use following fact:
\begin{align*}
  0<c_1(u)&\le \inf_{\lambda>0}\frac{\log\Big(1+\dfrac{u}{\lambda}\Big)}{\log\Big(1+\dfrac1{\lambda}\Big)}\\
&\le \sup_{\lambda>0}\frac{\log\Big(1+\dfrac{u}{\lambda}\Big)}{\log\Big(1+\dfrac1{\lambda}\Big)}
\le C_2(u)<\infty,\qquad \forall u>0.
\end{align*}
The proofs of sufficiency of (2),(3)  are direct.
Next the necessity discussed only.
In the following we always assume $\mu_n\overset{w}{\to}\mu^\ast$ and (6) holds.
A. If $ \psi^\ast(u)\ne0 $, then
\begin{equation*}
 \varliminf_{n\to\infty} \alpha_n>0,\qquad \varliminf_{n\to\infty} \lambda_n>0. \tag{8}
\end{equation*}
\textit{Proof}:
Prove (8) using contradictory. If (8) is not true, then there exists a subsequence
$ \alpha_{n_k}(\text{or }\lambda_{n_k})\to 0 $. From (6),
\begin{gather*}
 \lambda_{n_k}=\frac{\psi_{n_k}(u)}{e^{\psi_{n_k}(u)/\alpha_{n_k}}-1}\to0.\\
 \Big(\text{or } \alpha_{n_k}=\frac{\psi_{n_k}(u)}{\log\Big(1+\dfrac{u}{\lambda_{n_k}}\Big)}\to0.\Big)
\end{gather*}
Hence both $\alpha_{n_k}\to 0$, $\lambda_{n_k}\to 0$. If $\psi^\ast(a)\ne 0$,
\begin{equation*}
  \lim_{k\to\infty} \alpha_{n_k}\log\Big(1+\frac{a}{\lambda_{n_k}}\Big)
  =\lim_{k\to\infty} \psi_{n_k}(a)=\psi^{\ast}(a)\ne0.    \tag{9}
\end{equation*}
Now take $ u_{n_k}=\sqrt{\lambda_{n_k}^2+a\lambda_{n_k}}-\lambda_{n_k} $,
then $ u_{n_k}\to0 $ and
\begin{align*}
 \psi_{n_k}(u_{n_k})&=\alpha_{n_k}\log\Big(1+\frac{u_{n_k}}{\lambda_{n_k}}\Big)\\
 &=\frac12\alpha_{n_k}\log\Big(1+\frac{a}{\lambda_{n_k}}\Big)
 \to \frac12\psi^\ast(a)\ne0 \tag{10}
\end{align*}
(10) contradicts with (7). Hence (8) holds.
B. If $ \psi^\ast(u)\ne0 $ and there exists a subsequence  $\alpha_{n_k}\to\infty$(or $ \lambda_{n_k}\to\infty $ ), then
\begin{gather*}
 \lim_{n\to\infty}\alpha_n=\lim_{n\to\infty}\lambda_n=\infty,\\
 \lim_{n\to\infty}\frac{\alpha_n}{\lambda_n}=c, \qquad \psi^\ast(u)=cu.
\end{gather*}
\textit{Proof}: From (6), if $\alpha_{n_k}\to\infty$(or $ \lambda_{n_k}\to\infty $ ), then
\begin{equation*}
 \lambda_{n_k}=\frac{\psi_{n_k}(u)}{e^{\psi_{n_k}(u)/\alpha_{n_k}}-1}\to\infty,\quad
 \Big(\text{or } \alpha_{n_k}=\frac{\psi_{n_k}(u)}{\log\Big(1+\dfrac{u}{\lambda_{n_k}}\Big)} \to\infty\Big)
\end{equation*}
and
\begin{align*}
 \lim_{k\to\infty}\frac{\alpha_{n_k}}{\lambda_{n_k}}
 &=\lim_{k\to\infty}\frac{\psi_{n_k}(u)}{\lambda_{n_k}\log\Big(1+\dfrac{u}{\lambda_{n_k}}\Big)}\\
 &=\frac{\psi^\ast(u)}{u}\ne0, 
\end{align*}
$$ \psi^\ast(u)=cu.$$
In this case, if there exists a convergent subsequence $ \{\alpha_{n_k},k\ge 1\} $ (or $\{\lambda_{n_k},k\ge 1\}$), but not converge to $ \infty $, then the  $\psi_{n_k}(u)\to\psi^{\ast\ast}(u)\ne cu=\psi^\ast(u) $, so doesn't exist any subsequence $ \{\alpha_{n_k},k\ge 1\} $ (nor $\{\lambda_{n_k},k\ge 1\} $) which converge to a finite limit. B holds.
C. If $ \psi^\ast(u)\ne cu(c\ge 0) $, then
\begin{gather*}
  \lim_{n\to\infty}\alpha_n=\alpha^\ast\in(0,\infty),\\
\lim_{n\to\infty}\lambda_n=\lambda^\ast\in(0,\infty).\\
\psi^\ast(u)=\alpha^\ast\log\Big(1+\frac{u}{\lambda^\ast}\Big). 
\end{gather*}
\textit{Proof}: From {\bfseries B, C},  if $ \psi^\ast(u)\ne cu(c\ge 0) $, then
\begin{gather*}
 0<\varliminf_{n\to\infty} \alpha_n\le \varlimsup_{n\to\infty} \alpha_n<\infty,\\
 0<\varliminf_{n\to\infty} \lambda_n\le \varlimsup_{n\to\infty} \lambda_n<\infty.
\end{gather*}
This means that $S=\{(\alpha_n,\lambda_n), n\ge 1\} $ is a relative compact set in  $(0,+\infty)\times(0,+\infty)$(The closure $\overline{S}\subset(0,+\infty)\times(0,+\infty)$ and $\overline{S}$ is compact). Now we prove that $ A $ has unique limit point.
Suppose that  $\{(\alpha_{n_k},\lambda_{n_k}), k\ge 1\} $ is a convergent subsequence
in $ S $ and
\begin{align*}
 \alpha^\ast=\lim_{k\to\infty}\alpha_{n_k},\qquad \lambda^\ast=\lim_{k\to\infty}\lambda_{n_k}
\end{align*}
then
$\mu^*\overset{\text{d}}{=}\lim\limits_{k\to\infty}\mu_{n_k}$ is a $ \Gamma(\alpha^\ast,
\lambda^\ast) $-distribution and
\begin{equation*}
 \alpha^\ast=\frac{(\mathsf{E}_{\mu^\ast}[X])^2}{\mathsf{V}_{\mu^\ast}[X]},\quad
 \lambda^\ast=\frac{\mathsf{E}_{\mu^\ast}[X]}{\mathsf{V}_{\mu^\ast}[X]}.
\end{equation*}
This also means that for $ S $ there exits unique limits point $(\alpha^\ast,\lambda^\ast)$ and $ \mu^\ast\overset{\text{d}}{=}\Gamma(\alpha^\ast,\lambda^\ast) $.  C holds.
From above facts, it is easy to get the necessary conditions for weak convergence of Gamma-distribution.
A: In this comment I will give a full answer to the question about convergence of $X_n$.
We will say that the sequence $(\alpha_n, \lambda_n)$ is "nice" if it has the next property.
If for some subsequence $\alpha_{n_k}$ we have $\alpha_{n_k} \to c \in [0, \infty]$ then:

*

*if $c=0$ then $\liminf_{k} \lambda_{n_k}^{\alpha_{n_k}} \ge 1$

*if $c>0$ then $\lim_{k} \frac{\alpha_{n_k}}{\lambda_{n_k}}=0$.

Theorem 3:
If $X_n \overset{d}{\to}  X$ then $X$ may be $constant \ge 0$ or $X \sim \Gamma(\lambda, \alpha)$.
$$\Bigl( X_n \overset{d}{\to} 0 \Bigr) \Longleftrightarrow  \Bigl( (\alpha_n, \lambda_n) \text{ is "nice"} \Bigr).$$
$$\Bigl(X_n \overset{d}{\to}  X = C = const>0 \Bigr) \Longleftrightarrow \Bigl( \alpha_n \to \infty \text{ and } \frac{\alpha_n}{\lambda_n} \to C \in (0, \infty)  \Bigr)$$
\begin{gather}
\Bigl(X_n \overset{d}{\to}  X  \sim \Gamma(\lambda, \alpha) \Bigr) \Longleftrightarrow
\Bigl( \lim_n \alpha_n = \alpha \in (0, \infty) \text{ and } \lim_n \alpha_n = \lambda \in (0, \infty) \Bigr)\nonumber
\end{gather}
Corollary 1.
$X_n$ is convergent iff one of the following conditions holds true:

*

*$(\alpha_n, \lambda_n)$ is "nice"

*$\Bigl( \alpha_n \to \infty \text{ and } \frac{\alpha_n}{\lambda_n} \to C \in (0, \infty)  \Bigr)$

*$\lim_n \alpha_n = \alpha \in (0, \infty) \text{ and } \lim_n \alpha_n = \lambda \in (0, \infty)$.

Proof.
At first let us present results of my $2$ previous comments in a concise way.
If $\alpha_n \ge c > 0$ and $\lambda_n \to 0$ then $X_n$ diverges.
If $\alpha_n \to 0$ and $\lambda_n \ge c > 0$ then $X_n \overset{w}{\to} 0$.
If $\alpha_n \to 0$ and $\lambda_n \to 0$ then
$$X_n \overset{w}{\to} \Longleftrightarrow X_n \overset{w}{\to} 0 \Longleftrightarrow \Bigl(\lim_{n \to \infty} \alpha_n \ln \lambda_n = 0.\Bigr)$$
If for some $0<c_1 \le c_2 < \infty$ we have $\alpha_n, \lambda_n  \in [c_1, c_2]$  then
$ \Bigl( X_n \overset{w}{\to} \Bigr) \Longleftrightarrow \Bigl(\exists \lim_n \lambda_n$ and  $\exists \lim_n \alpha_n \Bigr)$. In this case $X_n \overset{w}{\to} \Gamma(\lim_n \lambda_n, \lim_n \alpha_n)$.
If $\alpha_n \le c_2$ and $\lambda_n \to \infty$ then $X_n  \overset{w}{\to} 0$.
If $\alpha_n \to \infty$ and $\lambda_n  \le c_2$ then $X_n$ diverges.
If  $\alpha_n \to \infty$ and $\lambda_n \to \infty$ then
$$X_n  \overset{w}{\to} \Longleftrightarrow \exists \lim_n \frac{\alpha_n}{\lambda_n},$$
and if $X_n$ is convergent then $X_n  \overset{w}{\to} \lim_n \frac{\alpha_n}{\lambda_n}$.
It's easy to see that if $X_n \overset{d}{\to}  X$ then $X$ may be $constant \ge 0$ or $X \sim \Gamma(\lambda, \alpha)$.
Moreover, it follows immediately that
$$\Bigl(X_n \overset{d}{\to}  X = C = const>0 \Bigr) \Longleftrightarrow \Bigl( \alpha_n \to \infty \text{ and } \frac{\alpha_n}{\lambda_n} \to C  \Bigr)$$
and
\begin{gather}
\Bigl(X_n \overset{d}{\to}  X  \sim \Gamma(\lambda, \alpha) \Bigr) \Longleftrightarrow 
\Bigl( \lim_n \alpha_n = \alpha \in (0, \infty) \text{ and } \lim_n \alpha_n = \lambda \in (0, \infty) \Bigr)\nonumber
\end{gather}
These conditions were mentioned here by JGWang.
The most interesting case is case when $X_n \overset{d}{\to}  0$.
We got convergence $X_n \overset{d}{\to}  0$ only in these cases:

*

*$\alpha_n \to 0$ and $\lambda_n \ge c > 0$,


*$\alpha_n \to 0$, $\lambda_n \to 0$ and $\lim_{n \to \infty} \alpha_n \ln \lambda_n = 0$,


*$\alpha_n \le c_2$, $\lambda_n \to \infty$,


*$\alpha_n \to \infty$, $\lambda_n \to \infty$ and $\lim_n \frac{\alpha_n}{\lambda_n}=0$.
These cases can be summarized as follows:

*

*$\alpha_n \to 0$ and $\Bigl( \limsup_n (-\alpha_n \ln \lambda_n) \le 0 \Longleftrightarrow \liminf_n \lambda_n^{\alpha_n} \ge 1 \Bigr)$

*$0 \le c_1 \le \alpha \le c_2 < \infty$ and $\lambda_n \to \infty$

*$\alpha_n \to \infty$ and $\lim_n \frac{\alpha_n}{\lambda_n}=0$.

Hence $X_n \overset{d}{\to} 0$ iff $(\alpha_n, \lambda_n)$ is "nice", q.e.d.
