Show that ${\sum\limits_{k=1}^n\frac{X_k}{\sigma_k^2}}\big/{\sum\limits_{k=1}^n\frac{1}{\sigma_k^2}}\overset{a.s.}\longrightarrow c.$ Let $X_1,X_2,\dots$ be independent random variables with $\lim\limits_{n\to\infty}\mathbb E[X_n]=c\in\mathbb R$ and $\sum\limits_{n=1}^\infty \frac1{\sigma_n^2}=\infty$.

I need to show that as $n\to\infty$
$${\sum\limits_{k=1}^n\frac{X_k}{\sigma_k^2}}\big/{\sum\limits_{k=1}^n\frac{1}{\sigma_k^2}}\overset{a.s.}\longrightarrow c.$$

I know that $\mathbb V\left[\frac{X_k}{\sigma_k^2}\right] = \frac{\mathbb V[X_k]}{\sigma_k^4} = \frac{1}{\sigma_k^2}$ from which follows with independence and the condition above that $\sum\limits_{k=1}^\infty\mathbb V\left[\frac{X_k}{\sigma_k^2}\right] = \infty$. How can I continue?
 A: Observe that the variance goes to $0$ :
\begin{align*}
\mathbb V\left[ \frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n \frac{1}{\sigma_k^2}} \right] &= \frac{\sum_{k=1}^n \frac{\sigma_k^2}{\sigma_k^4}}{\left(\sum_{k=1}^n \frac{1}{\sigma_k^2}\right)^2}\\
&=\frac{1}{\sum_{k=1}^n \frac{1}{\sigma_k^2}}\\
&\to 0
\end{align*}
and therefore
\begin{align*}
\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n \frac{1}{\sigma_k^2}}\overset{a.s.}\longrightarrow& \mathbb E\left[ \frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n \frac{1}{\sigma_k^2}} \right]\\
=&\frac{\sum_{k=1}^n \frac{c}{\sigma_k^2}}{\sum_{k=1}^n \frac{1}{\sigma_k^2}}\\
=&c
\end{align*}
A: By Chebyshev Inequality,
$$P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2  > \epsilon^2 \right) \leq \frac{Var\left( \frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} \right) }{\epsilon^2}\leq \frac{\left( \frac{\sum_{k=1}^n \frac{1}{\sigma_k^2}}{\left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)^2} \right) }{\epsilon^2} \leq \frac{1}{\epsilon^2 \left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)}$$
$$\lim_{n \rightarrow \infty} P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2  > \epsilon^2 \right) \leq \lim_{n \rightarrow \infty} \frac{1}{\epsilon^2 \left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)} = 0$$
If $$\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}}$$ exists then,
For all, $ n \geq N_{\delta,\epsilon,m}$ with $\lim_{m \rightarrow \infty} N_{\delta,\epsilon,m} = \infty$ such that
$$ P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2  > \epsilon^2 \right) \leq \frac{\delta}{2^{m}} \implies$$
$$ P \left( \cup_{m=1}^{\infty} \left \{ \left|\frac{\sum_{k=1}^{N_{\delta,\epsilon,m}} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{N_{\delta,\epsilon,m}}\frac{1}{\sigma_k^2}} -c\right|^2  > \epsilon^2 \right \} \right) \leq \delta \implies$$
$$ P \left( \left \{ \left|\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}} -c\right|^2  > \epsilon^2 \right \} \right) \leq \delta \implies$$
Since $\delta$ is arbitrary,
$$ P \left( \left \{ \left|\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}} -c\right|^2  > \epsilon^2 \right \} \right)  = 0$$
