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I wanted to complete the following sum: $$\sum_{0\leq i <j<k} a_ib_jc_k$$ Where $a_ib_jc_k$ are all different geometric sequences with $|r|<1$. My attempt was to break up the sum into what each thing is bounded by: $$\sum_{0\leq i <j<k} a_ib_jc_k=\sum_{i=0}^{j-1}\sum_{j=i+1}^{k-1}\sum_{k=j+1}^{\infty} a_ib_jc_k = \sum_{k=j+1}^{\infty}\sum_{j=i+1}^{k-1}\sum_{i=0}^{j-1}a_ib_jc_k$$ $$=\sum_{k=j+1}^{\infty}(\sum_{j=i+1}^{k-1}(\sum_{i=0}^{j-1}a_ib_jc_k)) \tag{1}$$

But I've also seen on stack exchange a discrete case, which if I extended it to this situation would make me believe the solution would be: $$\sum_{k=j+1}^{\infty}(\sum_{j=i+1}^{\infty}(\sum_{i=0}^{\infty}a_ib_jc_k))\tag{2}$$

I'm not sure if these are equivalent nor that either are correct for this case. For practical purposes, 1 would suggest that I compute $a_i$ as a finite geometric sum, making the summand become $a_ib_jc_k \to d_jc_k$ and then again to $f_k$ and I sum that to infinity subtracting the sum from 0 to base index for k which then leaves a result still in terms of the index k (which seems wrong).

And 2 would suggest that I could do something funky like separating out each series, evaluating them independently and then getting something weird where I need to subtract the sum from zero to the base index for each one: $$\sum_{k=j+1}^{\infty}(\sum_{j=i+1}^{\infty}(\sum_{i=0}^{\infty}a_ib_jc_k))=\sum_{k=j+1}^{\infty}c_k\sum_{j=i+1}^{\infty}b_j\sum_{i=0}^{\infty}a_i$$

What I think will work: if I realise that i is bounded below by 0 and j,k by 1 (consequently), then the sum over all i,j,k should be: $$\sum_{k=2}^{\infty}(\sum_{j=1}^{\infty}(\sum_{i=0}^{\infty}a_ib_jc_k))=\sum_{k=2}^{\infty}c_k\sum_{j=1}^{\infty}b_j\sum_{i=0}^{\infty}a_i=(\frac{1}{1-c}-1-c)(\frac{1}{1-b}-1)(\frac{1}{1-a})\tag{3}$$ Is (3) correct?

beigespectacles
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1 Answers1

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I do not think your (3) will work. To see this, $i<j<k$ is more severe restriction than $0\leq i, 1\leq j, 2\leq k$. Also (1) will not work, due to incorrect bounds. Actually, (2) is the correct approach, but with reverse order of summations (we treat inner sum first).

First, it is enough to evaluate the sum for $$\sum_{0\leq i<j<k} a^i b^j c^k.$$ The sum over $k$ is $k\geq j+1$, so it is $\frac{c^{j+1}}{1-c}$. Then take $c/(1-c)$ out of the sums, we are left with $$ \frac c{1-c}\sum_{0\leq i<j} a^i (bc)^j. $$

Then the sum over $j\geq i+1$ gives $$ \frac c{1-c} \frac{bc}{1-bc}\sum_{0\leq i} (abc)^i. $$ Thus, we have $$ \frac c{1-c} \frac{bc}{1-bc} \frac{1}{1-abc}. $$

Sungjin Kim
  • 20,102