# Is this sum convergent? [closed]

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

## closed as off-topic by heropup, José Carlos Santos, Leucippus, YuiTo Cheng, Ernie060Jun 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.

• How to justify that transformation? – Grešnik Jun 11 at 21:01
• @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. – Marian G. Jun 11 at 21:03
• I am asking how to justify the left hand side, not the right hand one. – Grešnik Jun 11 at 21:04
• @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. – lastresort Jun 12 at 5:58

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 $$3\leq i then you can take the cases when $$i=3,j=4$$ and get:

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

• Would anything change if we supposed that $i<j<k$? – Grešnik Jun 11 at 20:59
• Nope. Added that to my answer. @AnteP. – Thomas Andrews Jun 11 at 20:59

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.