Finding value of a summation series.

Question

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)?

Answer 1

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 $$ \frac{27}{8} - 3 \cdot\frac{27}{16} + 2 \frac{27}{26} \approx 0.38942\dots $$

Answer 2

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

$$\sum^{\infty}_{k=0}{1\over3^k}=\frac1{1-\frac13}=\frac32$$

$$\sum^{\infty}_{k=0}{1\over9^k}=\frac1{1-\frac19}=\frac98$$

$$\sum^{\infty}_{k=0}{1\over{27}^k}=\frac1{1-\frac1{27}}=\frac{27}{26}$$

Answer 3

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*}