$\{X_k\}_{k=1}^{\infty}$ are independent and identically distributed (the standard normal distribution),
$$Y_n^2=\sum_{k=1}^n X_k^2, \ \ \ \tau_n=\frac{X_{n+1}}{\sqrt{Y_n^2/n}}.$$
What is the distribution of $\lim_{n \longrightarrow \infty}\tau_n$?
$X^2_k$ are iid (squaring doesn't ruin it), and $EX_k^2=Var(X_k) + (EX_k)^2=1+0^2=1<\infty$, so by the law of large numbers $$ \frac{X_1^2+\ldots+X_n^2}{n} \longrightarrow EX_1^2=1 $$ in probability.
If I understand correctly, then $\sqrt{\frac{Y_n^2}{n}} \longrightarrow \sqrt{1} = 1$ in probability.
I can't write $\lim_{n \longrightarrow \infty}\tau_n=X_{n+1},$ and say that the answer is the standard normal distribution, since there can be no $n$ in the right side if $n$ is a limit parameter.
Is what I wrote above correct?
Could someone please help me show that $\lim \tau_n$ has the standard normal distribtion?
Hint: Note that $X_{n+1}$ is independent of $Y_n$ and $X_{n+1}$ converges to standard normal distribution.
If $A_n \to A$ in distribution, $B_n \to B$ a.s and $A$ is independent of $B$, $A_n$ is independent of $B_n$ for each $n$ then $A_nB_n \to AB$ in distribution.