# Evaluation of $\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{i+j+k+l}}\;\;,$ Where $i\neq j \neq k\neq l$

Evaluation of following Infinite Geometric series.

$(a)\;\; \displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^{i+j}}\;\;,$ Where $i\neq j\;\;\;\;\;\;\;\;\;\; (b)\;\; \displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{3^{i+j+k}}\;\;,$ Where $i\neq j \neq k$

$\displaystyle (c)\;\; \displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{i+j+k+l}}\;\;,$ Where $i\neq j \neq k\neq l.$

$\bf{My\; Try\; for }$ First one $(a)::$ Given $\displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^{i+j}} = \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^i}\cdot \frac{1}{3^j}\;\;,$ Where $i\neq j$

So we can write $\displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^i}\cdot \frac{1}{3^j}=\sum_{i=0}^{\infty}\frac{1}{3^i}\cdot \sum_{j=0}^{\infty}\frac{1}{3^j}-\sum_{i=0}^{\infty}\frac{1}{3^{2i}} = \frac{1}{1-\frac{1}{3}}\times \frac{1}{1-\frac{1}{3}}-\frac{1}{1-\frac{1}{3^2}}=\frac{9}{8}$

Actually i have used the fact $\displaystyle \sum(i\neq j) = \sum(\bf{no\; condition})-\sum(i=j)$.

But I did not understand how can i used the logic in $(b)$ and $(c)$ part , plz explain it to me in detail

Thanks

• I find $(b)$ and $(c)$ ambiguous: For instance, is $(b)$ excluding only indices like $(i,j,k)=(3,3,3)$, or are those like $(3,3,1)$ also forbidden? – Semiclassical Oct 31 '14 at 3:28

$$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{i+j+k+l}} \quad : i \neq j \neq k \neq l$$

The way the condition $i \neq j \neq k \neq l$ reads to me is $i \neq j \land j \neq k \land k \neq l$, but it could mean $\neg (i = j = k = l)$. We'll pursue both cases.

## Condition: $\neg (i = j = k = l)$

Since they're all the same, we can just use the $(\text{number of variables)} \times (\text{one of the variables})$ in place of $(i + j + k + l)$. This applies for the shorter cases too.

$$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{i+j+k+l}} - \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{4i}}$$

In this case, we chose $i$ and multiplied it by $4$.

## Condition: $(i, j, k, l)$ are pairwise-distinct

This one is easier to deal with because we're able to break this down properly with some choices.

$$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{i+j+k+l}} \\ - \underbrace{\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{4i}}}_\text{all variables are equal} \\ - \underbrace{2\binom{3}{1}\left(\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{2i + 2j}} - \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{4i}}\right)}_\text{two pairs of equal variables} \\ - \underbrace{\binom{4}{3}\left(\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{3i + j}} - \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{4i}}\right)}_\text{three variables are equal} \\ - \underbrace{\binom{4}{2}\left(\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{2i + j + k}} - \left(\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{3i + j}} - \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{4i}}\right) - \left(\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{2i + 2j}} - \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{4i}}\right)\right)}_\text{two variables are equal}$$

Edit: I removed one of the conditions because it was clearly not that one and I kept finding mistakes in it.

• OP may well also have meant the sum over all $(i, j, k, l)$ whose components are pairwise distinct, i.e., those tuples for which $i$, $j$, $k$, $l$ are all different. – Travis Willse Oct 31 '14 at 4:36
• I'll add that condition as well. – Axoren Oct 31 '14 at 4:38
• Your fortitude is admirable. :) – Travis Willse Oct 31 '14 at 4:39
• Thanks Axoren, would you like to explain me for $(b)$ part. – juantheron Oct 31 '14 at 4:41
• For part $(b)$:: Using $\displaystyle \sum_{i\ne j\ne k} =\sum_{i,j,k\ge0}-\sum_{i=j\ne k}-\sum_{i\ne j=k}-\sum_{i=k\ne j}-\sum_{i=j=k} =\sum_{i,j,k\ge0}-3\sum_{i=j\ne k}-\sum_{i=j=k}$ (By symmetry) $\displaystyle =\sum_{i,j,k\ge0}-3\left(\sum_{i=j}-\sum_{i=j=k}\right)-\sum_{i=j=k} =\sum_{i,j,k\ge0}-3\sum_{i=j}+2\sum_{i=j=k}.$ Is I am Right or not – juantheron Oct 31 '14 at 4:52

We have:

$$(a)\;\; S_2 = \sum_{i=0}^\infty\sum_{j=0}^\infty\frac{1}{3^{i+j}} (i \ne j) = 2\sum_{i< j}\frac{1}{3^{i+j}} = 2\;I_2$$

$$(b)\;\; S_3 = \sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty\frac{1}{3^{i+j+k}} (i \ne j \ne k) = 6\sum_{i<j<k}\frac{1}{3^{i+j+k}} = 6\;I_3$$

$$(c)\;\; S_4 = \sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty\sum_{l=0}^\infty\frac{1}{3^{i+j+k+l}} (i \ne j \ne k \ne l) = 24\sum_{i<j<k<l}\frac{1}{3^{i+j+k+l}} = 24\;I_4$$

And in general:

$$(d)\;\; S_n = \sum_{i_1=0}^\infty...\sum_{i_n=0}^\infty\frac{1}{3^{i_1+\cdots+i_n}} (i_1 \ne ... \ne i_n) = n!\sum_{i_1<...<i_n}\frac{1}{3^{i_1+\cdots+i_n}} = n!\;I_n$$

where

$$I_n = \sum_{i_1<...<i_n}\frac{1}{3^{i_1+\cdots+i_n}}$$

$I_1 = \sum_{i=0}^\infty 3^{-i} = \frac32$, and for $n \ge 1$:

\begin{align} I_{n+1} & = \sum_{i_1<...<i_{n+1}}\frac{1}{3^{i_1+\cdots+i_{n+1}}} \\ & = \sum_{i_1=0}^\infty \frac{1}{3^{i_1}} \sum_{i_1<i_2<...<i_{n+1}}\frac{1}{3^{i_2+\cdots+i_{n+1}}} \\ & = \sum_{i_1=0}^\infty \frac{1}{3^{i_1}} \sum_{0\le i_2<...<i_{n+1}}\frac{1}{3^{(i_1+i_2+1)+\cdots+(i_1+i_{n+1}+1)}} \\ & = \sum_{i_1=0}^\infty \frac{1}{3^{i_1}} \sum_{0 \le i_2<...<i_{n+1}}\frac{1}{3^{n(i_1+1)}}\frac{1}{3^{i_2+\cdots+i_{n+1}}}\\ & = \frac{1}{3^n}\sum_{i_1=0}^\infty \frac{1}{3^{(n+1)i_1}} \sum_{0 \le i_2<...<i_{n+1}}\frac{1}{3^{i_2+\cdots+i_{n+1}}}\\ & = \frac{1}{3^n}\frac{3^{n+1}}{3^{n+1}-1} I_n \\ & = \frac{3}{3^{n+1}-1}I_n \end{align}

Thus $S_{n+1} = (n+1)!\;I_{n+1} = \dfrac{3}{3^{n+1}-1}(n+1)!\;I_n = \dfrac{3(n+1)}{3^{n+1}-1}S_n$

Starting with $S_1 = I_1 = \frac32$, we get: $$S_2 = \frac{6}{8}S_1 = \frac{9}{8}$$ $$S_3 = \frac{9}{26}S_2 = \frac{81}{208}$$ $$S_4 = \frac{12}{80}S_3 = \frac{243}{4160}$$

• +1 I like this method alot because it doesn't require any combinatorial management. OP should consider this one for its cleaner logic as the accepted answer over mine. – Axoren Oct 31 '14 at 5:27
• I want to be your friend @Axoren – tmpys Oct 31 '14 at 9:48
• @juantheron: I have updated my answer to simplify the calculations. – TonyK Nov 1 '14 at 6:31