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The sum is: $$\sum_{3\leq i,j,k < + \infty} \frac {1}{ijk}$$.

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closed as off-topic by heropup, José Carlos Santos, Leucippus, YuiTo Cheng, Ernie060 Jun 12 at 8:14

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Your expression can be rewritten as

$$ \left (\sum_{i=3}^{\infty}\frac{1}{i}\right )\cdot\left (\sum_{j=3}^{\infty}\frac{1}{j}\right )\cdot\left (\sum_{k=3}^{\infty}\frac{1}{k}\right ) =\left (\sum_{i=3}^{\infty}\frac{1}{i}\right )^3. $$

Since the last sum is divergent (e.g., by using the integral criterion), your original sum is also divergent.

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    $\begingroup$ How to justify that transformation? $\endgroup$ – Grešnik Jun 11 at 21:01
  • $\begingroup$ @AnteP. The expression on the left is the product of three the same infinite sums. The variable notation is unimportant. Therefore, the right-hand side is in the form of a cube. $\endgroup$ – Marian G. Jun 11 at 21:03
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    $\begingroup$ I am asking how to justify the left hand side, not the right hand one. $\endgroup$ – Grešnik Jun 11 at 21:04
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    $\begingroup$ @AnteP. Since $i$, $j$ and $k$ are independent, we have $$\sum_{3 \le i, j, k \le \infty} \frac{1}{i j k} = \sum_{i=3}^\infty \sum_{j=3}^\infty \sum_{k=3}^\infty \frac{1}{i j k},$$ and the left hand side follows. $\endgroup$ – lastresort Jun 12 at 5:58
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No, because it include the summands when $i=j=3,$ which are $$\sum_{k=3}^{\infty}\frac{1}{9k},$$ which doesn't converge.

And if you want $3\leq i<j<k$ then you can take the cases when $i=3,j=4$ and get:

$$\sum_{k=5}^{\infty}\frac{1}{12k},$$

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  • $\begingroup$ Would anything change if we supposed that $i<j<k$? $\endgroup$ – Grešnik Jun 11 at 20:59
  • $\begingroup$ Nope. Added that to my answer. @AnteP. $\endgroup$ – Thomas Andrews Jun 11 at 20:59
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Clearly, $$\sum_{3\le i,j,k}\frac{1}{ijk}=\sum_{3\le k}\frac{1}{9k}+\sum_{3\le i-1,j-1,k}\frac{1}{ijk}$$ $$>\sum_{k\ge3}\frac{1}{9k}$$

and the Riemann series $\sum \frac 1k$ is divergent.

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