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?