A problem about the convergence of random sequence Let ($X_n,n\geq 1$) be a sequence of independent Guassian variables,with respective mean $\mu_n$ and variance ${\sigma_n}^2$.
(a)If $\Sigma_n{X_n^2}$ converges in $L^1$,then $\Sigma_n{X_n^2}$ converges in $L^p$,for every $p\in [1,\infty)$.
My solution is：Let $S_n=\Sigma_n{X_n^2}$ ,
$S_n\stackrel{L^1}{\longrightarrow}S$ and
$Y_n=-|{S_n-S}|^p$ ,
we have  $S_n\stackrel{P}{\longrightarrow}S$ and
$S_n-S\stackrel{P}{\longrightarrow}0$.
Now $Y_n\stackrel{P}{\longrightarrow}0$ since $f(x)=-|x|^p$ is continuous,then the Fatou lemma implies:
$0=E(\mathrm{lim}\,Y_n)\leq{-\mathrm{lim\, sup}_{n\rightarrow\infty}E(|{S_n-S}|^p})$.
I think that my solution is wrong,but don't know where is it.
(b)Assume that $\mu_n$=0,for every n.Prove that if $\Sigma_n{\sigma_n^2}=\infty$,then $\mathbb{P}(\Sigma_n{X_n^2}=\infty)=1.$
I had taken many methods to solute it,but all  failed.
 A: (a) Let $S_n = \sum_{k=1}^n X_k^2$. Since $L^p(\Omega,\mathcal F,\mathbb P)$ is a Banach Space, it is enough to prove that if $(S_n)_n$ is an $L^1-$Cauchy Sequence, then $(S_n)_n$ is $L^p-$Cauchy Sequence. Since $(S_n)_n$ is $L^1-$Cauchy Sequence, then for any $\varepsilon > 0$ there is some $N>0$ such that for any $m>n>N$ we have $\|S_n - S_m\|_{L_1} < \varepsilon $. But, $$ \|S_n-S_m\|_{L_1} = \mathbb E \left | \sum_{k=n+1}^m X_k^2 \right| = \mathbb E \left[ \sum_{k=n+1}^m X_k^2\right] = \sum_{k=n+1}^m \mathbb E[X_k^2] = \sum_{k=n+1}^m (\sigma_k^2 + \mu_k^2) < \varepsilon $$ means that (by Minkowsky-inequality (i.e - triangle inequality in $L_p$)) $$ \|S_n-S_m\|_{L_p} = \left( \mathbb E\left | \sum_{k=n+1}^m X_k^2 \right|^p \right)^{\frac{1}{p}} \le \sum_{k=n+1}^m \left( \mathbb E[ (X_k^2)^p] \right)^{\frac{1}{p}} $$ $$= \sum_{k=n+1}^m  \left( \mathbb E|\mu_k + \sigma_k\mathcal N(0,1)|^{2p} \right)^{\frac{1}{p}} < \sum_{k=n+1}^m \mu_k^2 + \sigma_k^2\left(\mathbb E|\mathcal N(0,1)|^{2p}\right)^{\frac{1}{p}}, $$ where we used the fact that $X_k$ has the same distribution as $\mu_k+ \sigma_k \cdot \mathcal N(0,1)$ (and the very last inequality is yet again Minkowsky inequality but in $L^{2p}$). In particular, $$ \|S_n - S_m\|_{L_p} \le \max\left\{1 , \left( \mathbb E|\mathcal N(0,1)|^{2p}\right)^{\frac{1}{p}}\right\} \sum_{k=n+1}^m \mu_k^2 + \sigma_k^2 < C_p \cdot \varepsilon,$$ which proves that $(S_n)_n$ is Cauchy in $L^p$.
(b) Firstly note that if $\sigma_n$ doesn't converge to $0$, then for some $\varepsilon > 0$ there are indices $\{n_k\}_k$ such that $\sigma_{n_k} > \varepsilon$. Let $G \sim \mathcal N(0,1)$, then $X_n \sim \sigma_n G$, so that $$ \sum_n\mathbb P(|X_n|^2 > 1) = \sum_n \mathbb P(|G| > \frac{1}{\sigma_n}) \ge \sum_k \mathbb P(|G|>\frac{1}{\sigma_{n_k}}) \ge \sum_k \mathbb P(|G| > \frac{1}{\varepsilon}) =\infty$$ hence by Borel Cantelli, $\sum_n X_n^2 = \infty$ almost surely. Assume now that $\sigma_n \to 0$. We have $$ \mathbb E[X_n^2 1_{|X_n| > 1}] = \sigma_n^2 \mathbb E[G^2 1_{\{|G| > \frac{1}{\sigma_n}\}}].$$ Since $|G|$ is square integrable and $\sigma_n \to 0$, we get $\mathbb E[G^2 1_{\{|G| > \frac{1}{\sigma_n}\}}] \to \mathbb E[G^2] = 1$. In particular, for $n$ big enough, $\mathbb E[X_n^2 1_{\{|X_n| > 1\}}] \ge \frac{\sigma_n^2}{2}$, which means that $$ \sum_{n=N} \mathbb E[X_n^2 1_{\{|X_n| > 1\}}] \ge \frac{1}{2}\sum_{n=N} \sigma_n^2 = \infty $$ so by Kołmogorov three series theorem, $\sum_{n=1}^\infty X_n^2$ diverges almost surely.
