Finding value of a summation series. 
Find the value of
  $$\sum^{\infty}_{i=0}\sum^{\infty}_{j=0}\sum^{\infty}_{k=0}{1\over3^i3^j3^k}$$
  Where $i\ne j,j\ne k,k\ne i$.  

What does this notion means? How can we write this in expanded form? What does the inequality between $i,j,k$ mean? And most importantly how to solve this type of series(hints would be better)?
 A: First ignore the restriction $i \ne j \ne k$. The result is $\left( \frac{1}{1 - \frac{1}{3}} \right)^3 = \frac{27}{8}$, since this is simply the geometric series $\sum_{i=0}^\infty \frac{1}{3^i}$ raised to the power 3.
Now subtract the three series that come from the conditions $i=j, i=k, j=k$. These are each equal to $\frac{1}{1 - \frac{1}{3}} \cdot \frac{1}{1 - \frac{1}{9}} = \frac{27}{16}$.
Finally add back two times the series coming from the terms with $i = j = k$, since it has been subtracted three times and should only be subtracted once. This series is equal to $\frac{1}{1 - \frac{1}{27}} = \frac{27}{26}$.
The result is 
$$
\frac{27}{8} - 3 \cdot\frac{27}{16} + 2 \frac{27}{26} \approx 0.38942\dots
$$   
A: Following the suggestion given in the comments we have that
$$\sum^{\infty}_{i=0}\sum^{\infty}_{j=0}\sum^{\infty}_{k=0}{1\over3^i3^j3^k}
=\left(\sum^{\infty}_{k=0}{1\over3^k}\right)^3
-3\sum^{\infty}_{k=0}{1\over3^k}\sum^{\infty}_{k=0}{1\over9^k}
+2\sum^{\infty}_{k=0}{1\over{27}^k}$$
and by geometric series
$$\sum^{\infty}_{k=0}{1\over3^k}=\frac1{1-\frac13}=\frac32$$
$$\sum^{\infty}_{k=0}{1\over9^k}=\frac1{1-\frac19}=\frac98$$
$$\sum^{\infty}_{k=0}{1\over{27}^k}=\frac1{1-\frac1{27}}=\frac{27}{26}$$
A: We can respect the conditions $i\ne j,j\ne k,k\ne i$ by adding up over $0\leq i<j<k<\infty$ and since the order counts we multiply with $3!$.

We obtain
  \begin{align*}
\color{blue}{3!}&\color{blue}{\sum_{0\leq i<j<k<\infty}\frac{1}{3^{i+j+k}}}\\
&=3!\sum_{i=0}^{\infty}\sum_{j=i+1}^{\infty}\sum_{k=j+1}^{\infty}\frac{1}{3^{i+j+k}}\\
&=3!\sum_{i=0}^{\infty}\sum_{j=i+1}^{\infty}\sum_{\color{blue}{k=0}}^{\infty}\frac{1}{3^{i+2j+k+1}}\tag{$k\to k+j+1$}\\
&=3!\sum_{i=0}^{\infty}\sum_{\color{blue}{j=0}}^{\infty}\sum_{k=0}^{\infty}\frac{1}{3^{3i+2j+k+3}}\tag{$j\to j+i+1$}\\
&=3!\cdot\frac{1}{1-\frac{1}{27}}\cdot\frac{1}{1-\frac{1}{9}}\cdot\frac{1}{1-\frac{1}{3}}\cdot\frac{1}{27}\\
&\,\,\color{blue}{=\frac{81}{208}\approx 0.38942}
\end{align*}

