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,
$$ P \left( \left \{ \left|\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right \} \right) = 0$$