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I asked here a number theory related question and got an answer. However, I am still having problems with understanding the following identity (which should hold for every $k\in \mathbb{N}_0$):

$$\prod_{p\in\mathbb{P}\cap\left[2{,}\ x\right]}^{ }\sum_{j=0}^k \frac{1}{p^j}=\sum_{i_2{,}\ i_3{,}\ \dots{,}\ i_n \in [0, \ k]}^{ }\frac{1}{2^{i_2}\cdot3^{i_3}\dots\cdot n^{i_n}}, $$

where $ \mathbb{P} \cap [2, x]=\left\{2, 3, \dots, n \right\} $ is the set of primes less than or equal to $x\ge 2$.

On some hand-waving level, I believe this statement, but I would like to see some rigorous reasoning (in math notation) which shows that LHS is the same as RHS. I tried with mathematical induction, but the induction step becomes too difficult. If the LHS is evaluated only "in one's mind" and using verbal description, one makes easily mistakes or wrong assumptions. That's why I would like more careful justifications.

In general, how to deal with this kind of product of sums (as on the LHS)?

mathslover
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2 Answers2

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Maybe writing the LHS in another way will help:

$$\Big(1 + \frac{1}{2} + \frac{1}{2^2} + \dots \frac{1}{2^k}\Big) \Big(1 + \frac{1}{3} + \frac{1}{3^2} + \dots \frac{1}{3^k}\Big)\dots \Big(1 + \frac{1}{n} + \frac{1}{n^2} + \dots \frac{1}{n^k}\Big)$$

To expand out these brackets, just pick a term from each bracket. For example $1$ appears in the expansion by picking the first term in each bracket and multiplying.

Can you make each term on the RHS by picking appropriate terms on the LHS (this would prove LHS $\geq$ RHS)? Given any choice of terms from each bracket on the LHS, when multiplied is it something of the form on the RHS (this would prove LHS $\leq$ RHS)?

Side note: $[0, k]$ is usually taken to mean $\{x: x \in \mathbb{R}, 0 \leq x \leq k\}$, I think you mean $[0, k] \cap \mathbb{Z}$.

You may also find this helpful: Rules of Double Sums

Tom Misch
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  • Answers are not discussions, I say this because you don't answer per se, you ask questions, which in turn requires a dialogue. But this site is not a discussion board. – amWhy Mar 29 '22 at 19:36
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I modified the notation slightly for convenience. You can achieve this by distributing the product of the primes from $p_2,...,.p_N$ to each of the $\dfrac{1}{p_1^j}$ (giving a sum on the outside). Then do the same for $p_2$, etc. Below $p_i$ is assumed to be the $n$th prime, $p$ a prime between $p_i$'s, and $j_i,k$ are all non-negative.

$$\prod_{p_1\le p\le p_N}\sum_{j\le k} \dfrac{1}{p^j}$$ $$=\sum_{j_1\le k}\left(\dfrac{1}{p_1^{j_1}}\prod_{p_2\le p\le p_N}\sum_{j\le k} \dfrac{1}{p^j}\right)$$ $$=\sum_{j_2\le k}\left(\dfrac{1}{p_2^{j_2}}\sum_{j_1\le k}\left(\dfrac{1}{p_1^{j_1}}\prod_{p_3\le p\le p_N}\sum_{j\le k} \dfrac{1}{p^j}\right)\right)$$ $$\vdots$$

Eventually you end up with an empty product in the 'middle', and so a nested sum of the form

$$\sum_{j_N\le k}\dfrac{1}{p_N^{j_N}}\left(\sum_{j_{N-1}\le k}\dfrac{1}{p_{N-1}^{j_{N-1}}}\left(\cdots\left(\sum_{j_1\le k}\dfrac{1}{p_1^{j_1}}\right)\right)\right)$$ From here, use distribution to obtain

$$\sum_{j_1\le k}\left(\sum_{j_2\le k}\left(\cdots \left(\sum_{j_N\le k}\dfrac{1}{p_1^{j_1}}\dfrac{1}{p_2^{j_2}}\cdots \dfrac{1}{p_N^{j_N}}\right)\right)\right)$$

The "$\cdots$" technically make this non-rigorous, but is really shorthand for simple induction.

David P
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