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][1]), 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 real numbers such that $\rho_n\to\infty$ and $\rho_n<<n/C_n^a$; 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. 


  [1]: http://www.amazon.com/dp/3642658113/?tag=stackoverfl08-20#reader_3642658113