Double sum, how to proceed? I'm struggling with the following problem. I get as variance
$$\operatorname{Var}(\hat{k})=C^2\sum_{i=0}^\infty \sum_{j=0}^\infty B^{i+j} \frac{\sigma_{\varepsilon}^2 \rho^{\,j-i}}{1-\rho^2}$$
Can I simplify this further? I tried
$$\operatorname{Var}(\hat{k})=\frac{C^2\sigma_{\varepsilon}^2}{1-\rho^2} \sum_{i=0}^\infty B^i \rho^{-i}\sum_{j=0}^\infty B^{\,j}\rho^{\,j}$$
and thus
$$\operatorname{Var}(\hat{k})=\frac{C^2\sigma_{\varepsilon}^2}{\left(1-\rho^2 \right) \left(1-\frac B \rho\right) (1-B\rho)}$$
but then I get a negative variance for the parameterization of the model. Moreover, I get $\operatorname{Var}(\hat{k})=0$ for $\rho=0$ but it is given that for $\rho=0$
$$\operatorname{Var}(\hat{k})=\frac{C^2\sigma_{\varepsilon}^2}{1-B^2}$$
Is it incorrect to "pull out" $B^i\rho^{-i}$ from the second into the first sum? Or is there any other mistake I don't see?
Can anybody help? I'd be really thankful guys!!!
edit
That's how I cam up with the first line. Given $\varepsilon_{t}\sim N(0,\sigma_{\varepsilon}^{2})$ are $i.i.d.$, $\hat{a}_{t+1}=\rho\hat{a}_{t} + \varepsilon_{t+1}$, the problem states then $ \hat{a}_{t+1} = \sum_{i=0}^{\infty}\rho^{i}\varepsilon_{t-i}$ as an approximation. 
I want to find
$$Var\left(C\sum_{i=0}^{\infty}B^i\hat{a}_{t-i}\right)$$
Thus,
\begin{split}
Var\left(C\sum_{i=0}^{\infty}B^i\hat{a}_{t-i}\right) &= Cov\left(C\sum_{i=0}^{\infty}B^i\hat{a}_{t-i},C\sum_{i=0}^{\infty}B^i\hat{a}_{t-i}\right) \\
&= C^2\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}B^{i+j}Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)
\end{split}
Then I used 
\begin{equation}
\begin{split}
Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right) &= Cov\left(\sum_{k=0}^{\infty}\rho^{k}\varepsilon_{t-i-k}, \sum_{l=0}^{\infty}\rho^{l}\varepsilon_{t-j-l}\right) \\
&= \sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\rho^{k+l}Cov\left(\varepsilon_{t-i-k},\varepsilon_{t-j-l}\right)
\end{split}
\end{equation}
Then I used that $Cov\left(\varepsilon_{t-i-k},\varepsilon_{t-j-l}\right)=0$ for $i+k \ne j+l$ so I put the double sum together in one sum with $k=j+l-i$
$$ Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)=\sum_{l=0}^{\infty}\rho^{2l+j-i}Cov\left(\varepsilon_{t-i-l},\varepsilon_{t-i-l}\right) = \sigma_{\varepsilon}^{2}\rho^{j-i}\sum_{l=0}^{\infty}\rho^{2l} = \frac{\sigma_{\varepsilon}^{2}\rho^{j-i}}{1-\rho^2}$$
and then I plugged this result in the double sum above to get
$$\operatorname{Var}(\hat{k})=C^2\sum_{i=0}^\infty \sum_{j=0}^\infty B^{i+j} \frac{\sigma_{\varepsilon}^2 \rho^{\,j-i}}{1-\rho^2}$$
Does anyone see the mistake??
 A: *

*For $\rho=0$ the starting sum will have a term $0^{j-i}$which tells
that you shall sum over $j=i$ thus getting 
$$\operatorname{Var}(\hat{k})=\frac{C^2\sigma_\varepsilon^2}{1-B^2}$$.
Then  $$ \eqalign{   & Var\left( {\hat k} \right) = {{C^{\,2} \sigma
_\varepsilon  ^{\,2} } \over {\left( {1 - \rho ^{\,2} } \right)}}\sum\limits_{0\, \le \,i} {\sum\limits_{0\, \le \,j}
{B^{\,j + i} \rho ^{\,j - i} } }  =   \cr    &  = \left\{ {\matrix{ 
{{{C^{\,2} \sigma _\varepsilon  ^{\,2} } \over {\left( {1 - B^{\,2}
} \right)}}} & {0 = \rho }  \cr     {{{C^{\,2} \sigma _\varepsilon 
^{\,2} } \over {\left( {1 - \rho ^{\,2} } \right)\left( {1 - B/\rho
} \right)\left( {1 - \rho B} \right)}}} & {0 < \left| \rho  \right|
< 1}  \cr   } } \right. \cr}  $$ which for certain values of $\rho$
and $B$ (e.g. $\rho=1/2,\,B=1$) gives negative value of the variance

*Concerning your addendum, regarding the recursion
$$
\hat a_{\,t + 1}  = \rho \hat a_{\,t}  + \varepsilon _{\,t + 1} \quad  \Rightarrow \quad \hat a_{\,t + 1\;\left( ? \right)}  = \sum\limits_{i = 0}^{\infty \;\left( ? \right)} {\rho ^{\,i} \varepsilon _{\,t - i} } 
$$
shouldn't the LHS be $a_{\,t}$ ? and the upper bound be $t$ ?
Also concerning
$$
Var\left( {C\sum\limits_{i = 0}^{\infty \;?} {B^{\,i} \hat a_{\,t - i} } } \right)
$$
the upper bound shall be $t$, otherwise you get negative indices for 
$a_{\,t-i}$

A: NEW SOLUTION PROPOSAL
Thanks for your answers guys. Now here's my new attempt with the remarks of @G Cab. 
\begin{equation}
\begin{split}
\label{eq:khat rho}
Var\left(\hat{k}_{t+1}\right) &= Var\left(C\sum_{i=0}^{t}B^i\hat{a}_{t-i}\right) = Cov\left(C\sum_{i=0}^{t}B^i\hat{a}_{t-i},C\sum_{i=0}^{t}B^i\hat{a}_{t-i}\right) \\
&= C^2\sum_{i=0}^{t}\sum_{j=0}^{t}B^{i+j}Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)
\end{split}
\end{equation}
Hence, we need $Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)$ as an intermediate result. Substituting for $\hat{a}_{t-i}$ and $\hat{a}_{t-j}$ in $Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)$ gives 
\begin{equation}
\begin{split}
\label{eq:cov1}
Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right) &= Cov\left(\sum_{k=0}^{t}\rho^{k}\varepsilon_{t-i-k}, \sum_{l=0}^{t}\rho^{l}\varepsilon_{t-j-l}\right) \\
&= \sum_{k=0}^{t}\sum_{l=0}^{t}\rho^{k+l}Cov\left(\varepsilon_{t-i-k},\varepsilon_{t-j-l}\right)
\end{split}
\end{equation}
Now note that $Cov\left(\varepsilon_{t-i-k},\varepsilon_{t-j-l}\right) = 0 \, \forall \, i+k \ne j+l$. Hence, put $k=j+l-i$ to reduce the above equation to
\begin{equation}
\label{eq:cov2}
Cov\left(\hat{a}_{t-i},\hat{a}_{t-j}\right)=\sum_{l=0}^{t}\rho^{2l+j-i}Cov\left(\varepsilon_{t-i-l},\varepsilon_{t-i-l}\right) = \sigma_{\varepsilon}^{2}\rho^{j-i}\sum_{l=0}^{t}\rho^{2l} = \frac{\sigma_{\varepsilon}^{2}\rho^{j-i}\left(1-\rho^{2(t+1)}\right)}{1-\rho^2}
\end{equation}
Then substituting this back in the original equation for the variance gives 
\begin{equation}
\begin{split}
Var\left(\hat{k}_{t+1}\right) &= C^2\sum_{i=0}^{t} \sum_{j=0}^{t} B^{i+j} \frac{\sigma_{\varepsilon}^{2}\rho^{j-i}\left(1-\rho^{2(t+1)}\right)}{1-\rho^2} \\
&= \frac{C^2\sigma_{\varepsilon}^2\left(1-\rho^{2(t+1)}\right)}{1-\rho^2} \sum_{i=0}^{t} B^i \rho^{-i}\sum_{j=0}^{t} B^{\,j}\rho^{\,j} \\
&= \frac{C^2\sigma_{\varepsilon}^2\left(1-\rho^{2(t+1)}\right)}{1-\rho^2}\sum_{i=0}^{t}\left(\frac{B}{\rho}\right)^{i}\sum_{j=0}^{t} \left(B\rho\right)^{j} \\
&= \frac{C^2\sigma_{\varepsilon}^2\left(1-\rho^{2(t+1)}\right)\left(1-\left(\frac{B}{\rho}\right)^{t+1}\right)\left(1-\left(B\rho\right)^{t+1}\right)}{\left(1-\rho^2\right)\left(1-\frac{B}{\rho}\right)\left(1-B\rho\right)} \\
\end{split}
\end{equation}
What do you think about this solution? Many thans in advance!!!
