The problem states that $\alpha>0$ and for each $n\in\mathbb{N},\ X_n:\Omega \rightarrow \mathbb{R}$ is a random variable on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with the gamma distribution $\Gamma_{\alpha,n}$. The question is if there exists a random variable $X:\Omega\rightarrow \mathbb{R}$ such that $X_n\overset{\mathbb{P}}{\to} X$ as $n\to\infty$.
Since our target here is quite clear that we want to show $$\forall\epsilon>0,\ \lim_{n\to\infty}\text{Pr}(\lvert X_n-X\rvert\leq\epsilon)=1$$
I'm not sure about the meaning of parameter $n$ here in a Gamma distribution. Is it a rate or a scale?
Suppose it is a rate, like $\Gamma_{\alpha,\beta}$, then in fact when $\beta=n\to\infty$,
$$\lim_{\beta\to\infty}X_\beta=\lim_{\beta\to\infty}\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}=\frac{x^{\alpha-1}}{\Gamma(\alpha)}\lim_{\beta\to\infty}\frac{\beta^\alpha}{e^{\beta x}}=0$$
Then we could just make it simple by setting $X$ to be a constant random variable, i.e. $\text{Pr}(X=0)=1.$ Hence such random variable exists.
Am I on the right track or missing something? Thank you.
You have mixed up random variables and their densities. $X_{\beta}$ is not equal to its density function.
According to the table in https://en.wikipedia.org/wiki/Gamma_distribution we have $EX_n=\frac {\alpha} n$.Since$EX_n \to 0$ and $X_n \geq 0$ it follows that $P(X_n >t)\leq \frac 1 t EX_n \to 0$ so $X_n \to 0$ in probability.