$X_i$ are $iid$, $EX_i = 0$, $EX_i ^2 = \sigma ^2$, $0<\sigma ^2<\infty$ Need to show:
$\frac{\sum_{m=1}^n X_m}{\sqrt{\sum_{m=1}^n X_m ^2}}\xrightarrow{w} N(0,1)$
My attempt:
Weak convergence $\iff$ convergence in characteristic functions
Characteristic func of RHS: $e^{-t^2 /2}$
Characteristic func of LHS: $E[\exp({it}\frac{\sum_{m=1}^n X_m}{\sqrt{\sum_{m=1}^n X_m ^2}})]$
$E[\exp({it}\frac{\sum_{m=1}^n X_m}{\sqrt{\sum_{m=1}^n X_m ^2}})] = \{E[1+\frac{\sum_{m=1}^{n}itX_m}{\sqrt{\sum_{m=1}^n X_m ^2}}-\frac{\sum_{m=1}^{n}t^2X_m^2}{2{\sum_{m=1}^n X_m ^2}}+o(t^2)]\}$ using Taylor's expansion
$\implies $ I need to show $E[\frac{itX_1}{\sqrt{\sum_{m=1}^n X_m ^2}}] = 0$ and $E[\frac{(\sum_{m=1}^n X_m)^2}{{\sum_{m=1}^n X_m ^2}}] = 1$, which I have trouble proving.
I can say $E[\frac{itX_1}{\sqrt{\sum_{m=1}^n X_m ^2}}] \leq E[\frac{itX_1}{\sqrt{\sum_{m=2}^n X_m ^2}}] = E[itX_1]E[\frac{1}{\sqrt{\sum_{m=2}^n X_m ^2}}] = 0$ since $EX_1 = 0$
but $X_i$ need not be positive and thus $E[\frac{itX_1}{\sqrt{\sum_{m=1}^n X_m ^2}}]$ need not be equal to zero.
A hint is appreciated. Thanks.
A simpler approach:
Write $$\frac{\sum_{m=1}^n X_m}{\sqrt{\sum_{m=1}^n X_m ^2}} = \frac{\sum_{m=1}^n X_m}{\sigma \sqrt{n}} \cdot \left(\frac{1}{\sigma^2} \cdot \frac{1}{n}\sum_{m=1}^n X_m ^2\right)^{-1/2}$$ Now use the central limit theorem, the law of large numbers, and Slutsky's theorem.