Find value of this Triple summation

Question

Find the value of $$S=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{3^{i+j+k}}$$ where $i \ne j \ne k$

i have tried in this way:

First fix $j$ and $k$, then we get

$$S_1=\sum_{\substack{i=0 \\ i\neq j,k}}^{\infty}\frac{1}{3^{i+j+k}}$$ where $j \ne k$

So we get $$S_1=\left(\frac{3}{2} \times\frac{1}{3^{j+k}}\right)-\frac{1}{3^{2j+k}}-\frac{1}{3^{j+2k}}$$

So now

$$S_2=\sum_{\substack{j=0 \\ j\neq k}}^{\infty}\left(\left(\frac{3}{2} \times\frac{1}{3^{j+k}}\right)-\frac{1}{3^{2j+k}}-\frac{1}{3^{j+2k}}\right)$$

So $$S_2=\left(\frac{9}{4} \times \frac{1}{3^k}\right)-\left(\frac{9}{8} \times \frac{1}{3^k}\right)-\left(\frac{3}{2}\times \frac{1}{9^k}\right)-\left(\frac{3}{2}\times \frac{1}{3^{2k}}\right)+\left(\frac{2}{3^{3k}}\right)$$

Finally

$$S=\sum_{k=0}^{\infty}\left(\frac{9}{4} \times \frac{1}{3^k}\right)-\left(\frac{9}{8} \times \frac{1}{3^k}\right)-\left(\frac{3}{2}\times \frac{1}{9^k}\right)+\sum_{k=0}^{\infty}\frac{2}{3^{3k}}-\sum_{k=0}^{\infty}\frac{\frac{3}{2}}{3^{2k}}$$

we get

$$S=\frac{27}{16}-\frac{27}{16}+\frac{54}{26}-\frac{27}{16}$$ hence

$$S=\frac{81}{208}$$

is there any other approach like using integration etc

Answer 1

You have $$S=S_0-S_1-S_2+S_3$$ where $S_0$ is the sum over all triples $(i,j,k)$, $S_1$ is the sum over triples $(i,j,k)$ with $i=j$, $S_2$ is the sum over triples $(i,j,k)$ with $j=k$ and $S_3$ is the sum over triples $(i,j,k)$ with $i=j=k$. This is basically inclusion-exclusion. Then $$S_0=\left(\frac32\right)^3=\frac{27}8,$$ $$S_1=S_2=\sum_{i=0}^\infty\frac1{9^i}\sum_{k=0}^\infty\frac1{3^k} =\frac{27}{16}$$ and $$S_3=\sum_{i=0}^\infty\frac1{27^i}=\frac{27}{26}$$ etc.

ADDED IN EDIT If you actually want the sum over $(i,j,k)$ with $i$, $j$, $k$ distinct, rather than with just $i\ne j\ne k$, then inclusion-exclusion will give the answer $S_0-3S_1+2S_3$ instead.

Answer 2

For all $i,j,k$,

$$A=\sum_{i=0}^\infty\frac 1{3^i}=\frac 1{1-\frac 13}=\frac 32\\ \therefore A^3=\frac {27}8$$ For $i=j(=r)\neq k$,

$$B=\sum_{k=0}^\infty\frac 1{3^k}\sum_{r=0}^\infty \frac 1{3^{2r}}=\frac 32\cdot \frac 1{1-\frac 19}=\frac 32\cdot \frac 98=\frac {27}{16}$$

For $i=j=k(=r)$,

$$C=\sum_{r=0}^\infty \frac 1{3^{3r}}=\frac 1{1-\frac 1{27}}=\frac {27}{26}$$ The required sum, $$\sum_{i,j,k=0\\i\neq j\neq k}^\infty\frac 1{3^{i+j+k}}=A^3-3B+2C=27\left(\frac 18-\frac 3{16}+\frac 1{13}\right)=\color{red}{\frac {81}{208}}$$