Evaluate the following triple summation 
Evaluate $$\sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \sum_{i=0}^{\infty} \frac{1}{3^i3^j3^k}$$ for $i\neq j\neq k$ 

My Attempt
Evaluate the summation with no restriction on $i,j,k$. Let this be $a_0$  
When $i= j=k$. Let this be $a_1$.  
When $i\neq j=k$, let this be $a_2$.  
The required answer would be $a_0-a_1-3a_2$.
I got \begin{align}a_0&=\frac{27}{8}\\\\
a_1&=\frac{27}{26}\\  \\
a_2&=\frac{135}{208}\end{align}  
And therefore, my answer is 
\begin{align}a_0-a_1-3a_2&=\frac{27}{8}-\frac{27}{26}-3\times\frac{135}{208}\\\\
&=\frac{81}{208}\end{align} 
Is everything correct? The method and the calculation? Also if anyone has a better method, please tell. 
 A: Let's rearrange the sum: 
$$ S := \sum_{i,j,k \ge 0, \ i \neq j \neq k} 3^{-i} 3^{-j} 3^{-k} = \sum_{j \ge 0} \left( 3^{-j} \sum_{i \ge 0, \ i \neq j} 3^{-i} \sum_{k \ge 0, \ k \neq j} 3^{-k} \right). $$
The two inner sums are equal 
$$ \sum_{i \ge 0, \ i \neq j} 3^{-i} = \left( \sum_{i \ge 0} 3^{-i} \right) - 3^j = \frac{3}{2} - 3^{-j}, $$
so we're left with 
\begin{align*}
S & = \sum_{j \ge 0} \left( 3^{-j} \left(\frac{3}{2} - 3^{-j}\right)^2 \right) \\
& = \left( \frac 32 \right)^2 \sum_{j \ge 0} 3^{-j} 
- 2 \cdot \frac 32 \cdot \sum_{j \ge 0} 3^{-2j}
+ \sum_{j \ge 0} 3^{-3j} \\
& = \frac 94 \cdot \frac 32 - 3 \cdot \frac 98 + \frac{27}{26} \\
& = \frac{27}{26}.
\end{align*}

If I understand your notation correctly, the sum should be equal to 
$$ S = a_0 - a_1 - 2 a_2, $$
but substituting the values of $a_0,a_1,a_2$ you calculated gives a negative result. Since you haven't included your reasoning, I cannot say more. 

If you're interested in $i,j,k$ pairwise distinct, this is a different problem. By the inclusion-exclusion principle, we get 
\begin{align*}
\sum_{i,j,k \text{ distinct}} 3^{-i-j-k}
& = \left( \sum_{i,j,k} 3^{-i-j-k} \right)
- 3 \left( \sum_{i=j, \ k} 3^{-i-j-k} \right) + 
2 \left( \sum_{i=j=k} 3^{-i-j-k} \right) \\ 
& = \left( \frac 32 \right)^3 - 3 \cdot \frac 32 \cdot \frac 98 + 2 \cdot \frac{27}{26} \\
& = \frac{81}{208},
\end{align*}
which happens to coincide with your result.
A: I have the same answer.
\begin{align*}
\sum_{k\in\mathbb{N}} \sum_{j\in\mathbb{N}\setminus\{k\}} \sum_{i\in\mathbb{N}\setminus\{j,k\}} \frac{1}{3^i3^j3^k}&=\sum_{k\in\mathbb{N}} \sum_{j\in\mathbb{N}\setminus\{k\}} \left(\frac{1}{3^j3^k}\sum_{i\in\mathbb{N}\setminus\{j,k\}} \frac{1}{3^i}\right)\\
&=\sum_{k\in\mathbb{N}} \sum_{j\in\mathbb{N}\setminus\{k\}} \left[\frac{1}{3^j3^k}\left(\frac{1}{1-\frac{1}{3}}-\frac{1}{3^j}-\frac{1}{3^k}\right)\right]\\
&=\sum_{k\in\mathbb{N}}\left[\frac{1}{3^k}\left(\frac{3}{2}-\frac{1}{3^k}\right)\sum_{j\in\mathbb{N}\setminus\{k\}} \frac{1}{3^j}-\frac{1}{3^k}\sum_{j\in\mathbb{N}\setminus\{k\}} \frac{1}{3^{2j}}\right]\\
&=\sum_{k\in\mathbb{N}}\left[\frac{1}{3^k}\left(\frac{3}{2}-\frac{1}{3^k}\right) \left(\frac{3}{2}-\frac{1}{3^k}\right)-\frac{1}{3^k}\left(\frac{1}{1-\frac{1}{9}}-\frac{1}{3^{2k}}\right)\right]\\
&=\sum_{k\in\mathbb{N}}\left(\frac{9}{8}\cdot\frac{1}{3^k}-3\cdot\frac{1}{3^{2k}}+2\cdot\frac{1}{3^{3k}}\right)\\
&=\frac{9}{8}\cdot\frac{3}{2}-3\cdot\frac{9}{8}+2\cdot\frac{1}{1-\frac{1}{27}}\\
&=\frac{9}{8}\cdot\frac{3}{2}-3\cdot\frac{9}{8}+2\cdot\frac{27}{26}\\
&=\frac{81}{208}
\end{align*}
