The law of large numbers for diverging moments I have a random variable $X$ whose first and second moments are given as
$$
E[X] \propto C_n^{1-a},\quad E[X^2] \propto C_n^{2-a}\quad (0 < a < 1),
$$
where $C_n$ satisfies
$$
\lim_{n\to\infty} C_n = \infty, \quad \lim_{n\to\infty} \frac{C_n^a}{n} = 0.
$$
I want to see the limit of a ratio, 
$$
R_n = \frac{\sum_{i=1}^n X_i^2}{(\sum_{i=1}^n X_i)^2},
$$
where $X_1,\dots, X_n$ is i.i.d. samples of $X$. If the first and second moment is finite, one can easily see that
$$
\lim_{n\to\infty} R_n = \lim_{n\to\infty} \frac{1}{n} \frac{\sum_{i=1}^n X_i^2/n}{(\sum_{i=1}^n X_i)^2/n^2} = 0,
$$
by the law of large numbers. But in this case, $C_n$ also goes to infinity and as far as I know, the law of large numbers does not hold for infinite moments. At first, I forgot this fact and just thought as follows,
$$
\lim_{n\to\infty} R_n \propto \lim_{n\to\infty} \frac{1}{n} \frac{C_n^{2-a}}{C_n^{2-2a}}
= \lim_{n\to\infty} \frac{C_n^a}{n} = 0,
$$
by the condition given to $C_n$. I empirically checked whether $R_n$ converges to zero as $n\to\infty$, and it indeed goes to zero. Is there any clear proof, or if it is wrong, can anybody explain why it is wrong?
Thanks in advance for the answers.

Sorry for the confusion. What Iosif Pinelis interpreted is right. To be more specific, I have a r.v. $X_n$ constrained on an interval $[0, C_n]$, and its first and second moments
are given as
$$
EX_n = \frac{C_n^{1-a}}{1-a} - \gamma(1-a, C_n),\quad EX_n^2 = \frac{C_n^{2-a}}{2-a} - \gamma(2-a, C_n),
$$
where $\gamma(\cdot,\cdot)$ is a lower incomplete gamma function. I compute the ratio $R_n$ by drawing $n$ i.i.d. samples $X_{n,1}, \dots X_{n,n}$ of $X_n$. For a fixed $n$ (and thus $C_n$), there is no problem, but I want to know how $R_n$ behaves as $n\to\infty$, so $C_n\to\infty$. 
 A: I interpret the question as follows (cf. the comment by Nate Eldredge).
For each natural $n$, let $X_n,X_{n,1},\dots,X_{n,n}$ be independent identically distributed (i.i.d.) random variables (r.v.'s) such that 
$$
EX_n \bowtie C_n^{1-a},\quad EX_n^2\bowtie C_n^{2-a}\quad (0 < a < 1),
$$
where $C_n$ satisfies
$$(1)\qquad
\lim_{n\to\infty} C_n = \infty, \quad \lim_{n\to\infty} \frac{C_n^a}{n} = 0, 
$$
and $a\bowtie b$ means that $a\triangleleft b\triangleleft a$ and $a\triangleleft b$ means $a=O(b)$. 
Let
$$
R_n := \frac{\sum_{i=1}^n X_{n,i}^2}{(\sum_{i=1}^n X_{n,i})^2}. 
$$
Show that $R_n\to0$; here and elsewhere, the convergence of r.v.'s is in probability, as $n\to\infty$.

The conclusion $R_n\to0$ is indeed true and can be proved as follows, actually under the more general condition
$$(2)\qquad 
EX_n \triangleright C_n^{1-a},\quad EX_n^2 \triangleleft C_n^{2-a}\quad (0 < a < 1),
$$
where $a\triangleright b$ means $b\triangleleft a$. 
For each natural $n$, let $V_n,V_{n,1},\dots,V_{n,n}$ be i.i.d. r.v.'s. 
By the well-known necessary and sufficient condition for the (weak) law of large numbers (LLN) (see e.g. Theorem 3 of Ch. IX of Petrov), for $\sum_{i=1}^n V_{n,i}\to0$ it is sufficient that for each real $\tau>0$ 
(i) $nP(|V_n|\ge\tau)\to0$, 
(ii) $nEV_n^2\,I\{|V_n|<\tau\}\to0$, and 
(iii) $nEV_n\,I\{|V_n|<\tau\}\to0$, where $I$ denotes the indicator.  
Let now $Y_{n,i}:=X_{n,i}^2/B_n$ and $Y_n:=X_n^2/B_n$, where $B_n:=\rho_n n C_n^{2-a}$ and the $\rho_n$'s are any positive real numbers such that $\rho_n\to\infty$ and $\rho_n C_n^a/n\to0$; by (1), such $\rho_n$'s exist. Then, by (2), for each real $\tau>0$,
$$nP(|Y_n|\ge\tau)\le nE|Y_n|/\tau=nEX_n^2/(\tau B_n)\triangleleft nC_n^{2-a}/B_n=1/\rho_n\to0,$$ 
$$nEY_n^2\,I\{|Y_n|<\tau\}\le\tau nE|Y_n|\triangleleft1/\rho_n\to0\ \text{(cf. the previous line)},$$
$$\big|nEY_n\,I\{|Y_n|<\tau\}\big|\le nE|Y_n|\triangleleft1/\rho_n\to0.$$ 
So, by the above sufficient condition for the LLN, 
$$(3)\qquad \sum_{i=1}^n X_{n,i}^2/B_n=\sum_{i=1}^n Y_{n,i}\to0.
$$
Next, let $Z_{n,i}:=(X_{n,i}-EX_{n,i})/E_n$ and $Z_n:=(X_n-EX_n)/E_n$, where $E_n:=nC_n^{1-a}$. Then, by (2) and (1), for each real $\tau>0$,
$$nP(|Z_n|\ge\tau)\le nEZ_n^2/\tau^2\triangleleft nEX_n^2/E_n^2\triangleleft nC_n^{2-a}/E_n^2=C_n^a/n\to0,
$$ 
$$nEZ_n^2\,I\{|Z_n|<\tau\}\le nEZ_n^2\to0\ \text{(cf. the previous line)},$$
$$\big|nEZ_n\,I\{|Z_n|<\tau\}\big|=\big|nEZ_n\,I\{|Z_n|\ge\tau\}\big|\le nEZ_n^2/\tau\to0.$$ 
So, by the same sufficient condition for the LLN, $\sum_{i=1}^n Z_{n,i}\to0$; that is, 
$\sum_{i=1}^n X_{n,i}/E_n-nEX_n/E_n\to0.
$
On the other hand, by (2), $nEX_n/E_n\triangleright1$. So, $\sum_{i=1}^n X_{n,i}/E_n\triangleright1$ and hence 
$$\Big(\sum_{i=1}^n X_{n,i}\Big)^2\triangleright E_n^2=n^2C_n^{2-2a}. 
$$
Comparing this with (3), one concludes that 
$$R_n<<\frac{B_n}{E_n^2}=\rho_n C_n^a/n\to0, 
$$
by the choice of $\rho_n$. 
This completes the proof. 
A: See theorem 1 in Erickson's 1973 paper.
