I have troubles applying the Law of Large Numbers (abbreviated LLN) when the random variable is transformed. Let all $X_i$ be independent and identically distributed and $X_i\in L^2(P)$.
For example I've read that when $$S_n=X_1^2+\cdots + X_n^2$$ Then the LLN says that $$\sum\limits_{n=1}^\infty\frac{S_n}{n}=\mathbb{E}(X^2)$$
Does that mean when $f\in L^1$ $$S_n=f(X_1)+\cdots + f(X_n)$$ Then the LLN says that $$\sum\limits_{n=1}^\infty\frac{S_n}{n}=\mathbb{E}(f(X))$$
Second example. If I have $f(S_n)$ like for instance $$\frac{1}{\sqrt{1-(n^{-1}S_n)^2}}=\frac{1}{\sqrt{1-(n^{-1}(X_1+\cdots+X_n))^2}}$$
Does that mean that the LLN say that $$\frac{1}{\sqrt{1-(n^{-1}S_n)^2}}\xrightarrow{\text{a.s.}} \frac{1}{\sqrt{1-(\mathbb{E}(X))^2}}$$
Thank you for explanation.
The LLN says that, if $Y_1,\ldots,Y_n,\ldots$ are independent and identically distributed random variables in $L^1$, then $$\frac{1}{n}\sum_{k=1}^nY_k\stackrel{\text{a.s.}}{\rightarrow}\mu,$$ where $\mu=E[Y_i]$ for all $i$.
If you call $S_n=Y_1+\ldots +Y_n$, then $$\frac{S_n}{n}\stackrel{\text{a.s.}}{\rightarrow}\mu$$ (your notation with the $\sum_{n=1}^{\infty}$ is not correct).
If $Y_i=f(X_i)$, you need $f(X_i)\in L^1$ to apply the LLN and conclude that $$\frac{1}{n}\sum_{k=1}^nf(X_k)\stackrel{\text{a.s.}}{\rightarrow}E[f(X)].$$
In your final question, $n^{-1}S_n\rightarrow E[X]$ by the LLN. Assuming that $n^{-1}S_n\in (-1,1)$ and $E[X]\in (-1,1)$, as the function $x\mapsto 1/\sqrt{1-x^2}$ is continuous on $(-1,1)$, you can apply the continuous mapping theorem: $$ \frac{1}{\sqrt{1-(n^{-1}S_n)^2}}\stackrel{\text{a.s.}}{\rightarrow} \frac{1}{\sqrt{1-E[X]^2}}.$$