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We can be sure, that $$(1-a_{1}x)(1-a_{2}x)(1-a_{3}x)\cdots=1-(a_{1}+a_{2}+a_{3}+\cdots)x+(a_{1}a_{2}+a_{1}a_{3}+\cdots)x^2-(a_{1}a_{2}a_{3}+\cdots)x^3+\cdots$$ Other words, $$\prod\limits_{k=1}^{\infty}(1-a_{k}x)=1-A_{1}x+A_{2}x^2-A_{3}x^3+\cdots$$ Let create a function $$B(n)=\sum\limits_{k=1}^{\infty}a_{k}^{n}$$ Then we can say, that $$B(1)=A_{1}, B(2)=B(1)A_{1}-2A_{2}=A_{1}^{2}-2A_{2}$$ $$B(3)=B(2)A_{1}-B(1)A_{2}+3A_{3}=A_{1}^{3}-3A_{1}A_{2}+3A_{3}$$ $$B(4)=B(3)A_{1}-B(2)A_{2}+B(1)A_{3}-4A_{4}=A_{1}^{4}-4A_{1}^{2}A_{2}+4A_{1}A_{3}+2A_{2}^{2}-4A_{4}$$ In general $$B(n)=(-1)^{n-1}nA_{n}+\sum\limits_{k=1}^{n-1}(-1)^{k-1}B(n-k)A_{k}$$ So if we create a function $$C(n)=(a_{1}a_{2})^n+(a_{1}a_{3})^n+(a_{2}a_{3})^n+\cdots$$ or $$D(n)=(a_{1}a_{2}a_{3})^{n}+\cdots$$ which uses infinite sums of $A_{2}$ and $A_{3}$, how can we find it in general (also not only for this two, but for any $A_{m}$)?

If I made some mistakes, sorry for my English.

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  • $\begingroup$ I am not quite sure what you are asking here. Didn't you just describe the coefficients yourself? $\endgroup$ – Fimpellizieri Jan 29 '18 at 19:19
  • $\begingroup$ Can you be sure that each of the $A_i$'s converge? $\endgroup$ – Doughnut Pump Jan 29 '18 at 19:20
  • $\begingroup$ @DoughnutPump Is this even relevant? $\endgroup$ – Fimpellizieri Jan 29 '18 at 19:22
  • $\begingroup$ @Fimpellizieri, thank you for answer! I describe only one case, which use $A_{1}$, and interested in others. $\endgroup$ – user514787 Jan 29 '18 at 19:25
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    $\begingroup$ Is your question about 'How do I write these expressions in a compact form?' $\endgroup$ – Fimpellizieri Jan 29 '18 at 19:34
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If $S_{k,n}$ is the sum of the $n$-th powers of products of $k$ terms $a_i$, then you can write

$$S_{k,n}=\sum_{1\leq i_1<i_2<\ldots<i_k} {\left(a_{i_1}\cdot a_{i_2}\cdot \ldots\cdot a_{i_k}\right)}^n$$


Write

$$f_m(x)=\prod_{k=1}^\infty(1-a_k^mx)$$

and let $A_{k}(m)$ be the coefficient of $[x^k]\Big(f_m(x)\Big)$. So, using your notation we'd have $A_j=A_j(1)$.

Let $B_n(m)=\sum_{k\geq 1}a_k^{mn}$. So,using your notation, we'd have $B(n)=B_1(n)=B_n(1)$.

Observe that the relations you found between $B$ and $A$ are valid in greater generality with

$$B_n(m)=(-1)^{n-1}nA_{n}(m)+\sum\limits_{k=1}^{n-1}(-1)^{k-1}B_{n-k}(m)A_{k}(m)$$

Moreover, note that in your notation $C(n)=A_2(n)$ and $D(n)=A_3(n)$, and so on.

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  • $\begingroup$ Sorry, now I think you don't understand me right. I need form like $B(n)$ after words "In general" for $C(n)$, $D(n)$ and any other cases. $\endgroup$ – user514787 Jan 29 '18 at 19:42
  • $\begingroup$ See the edit. I still find your question much too vague, but eh.] $\endgroup$ – Fimpellizieri Jan 29 '18 at 20:07

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