Dividing by $\sum X_i^2$ rather than its square root leaves a $N(0,1/\sigma^2)$ limit

Suppose $\{X_1,X_2,\dots,X_n\}$ are independent and identically distributed random variables with mean zero and variance $\sigma^2$. Then the asymptotic distribution of

$$T_n=\frac{\sqrt n\sum_{i=1}^nX_i}{\sum_{i=1}^nX_i^2}$$

is: