I am trying to prove the equation

\begin{align*} \sum_{n=0}^{\infty} a^n \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d} = \prod_{i=1}^d \frac{1}{1 - a b_i}, \end{align*}

where we have that $n = c_1 + c_2 + ... c_d$. This is what I have tried so far:

\begin{align*} \sum_{n=0}^{\infty} a^{c_1 + c_2 + \ldots + c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d} &= \sum_{n=0}^{\infty} a^{c_1} \cdot a^{c_2} \cdot \cdot \cdot a^{c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdot \cdot \cdot b_d^{c_d} \\ &= \sum_{c_1}^{\infty} a^{c_1} \cdots \sum_{c_d}^{\infty} a^{c_d} \cdot \sum_{c_1}^{} b_1^{c_1} \cdots \sum_{c_d} b_d^{c_d} \\ &= \sum_{c_1}^{\infty} (a b_1)^{c_1} \cdots \sum_{c_d}^{\infty} (ab_d)^{c_d}. \end{align*}

My doubts in here are whether it is even possible to split up the sum in this manner and then simply recombining it in the last line, since the product of two sums is not necessarily the sum of the product.

From here I would use

\begin{align*} \sum_{k}^{\infty} x^k = \frac{1}{1-x}, \end{align*}

such that it would become

\begin{align} \frac{1}{1-a b_1} \times \cdots \times\frac{1}{1 - a b_d} = \prod_{i = 1}^{d} \frac{1}{1-a b_i}. \end{align}

I would like to ask where my errors are and if there is a better way to prove this identity. Thank you for your time.


2 Answers 2


By induction on $d$.

$$\begin{aligned} \frac{1}{1-ab_1} \frac{1}{1-ab_2} &= \left(\sum_{n=0}^\infty (ab_1)^n\right)\left(\sum_{n=0}^\infty (ab_2)^n\right)\\ &=\sum_{n=0}^\infty \left(\sum_{l=0}^nb_1^lb_2^{n-l}\right) a^n\\ &=\sum_{n=0}^\infty \left(\sum_{c_1+c_2=n}b_1^{c_1} b_2^{c_2}\right) a^n \end{aligned}$$

according to Cauchy product formula.

Now apply again Cauchy product formula to pass from $d$ to $d+1$:

$$\begin{aligned} \prod_{i=1}^{d+1} \frac{1}{1 - a b_i} &=\left(\prod_{i=1}^{d} \frac{1}{1 - a b_i}\right)\frac{1}{1 - a b_{d+1}}\\ &= \left(\sum_{n=0}^\infty \left(\sum_{c_1+ \dots +c_d=n}b_1^{c_1} \dots b_d^{c_d}\right) a^n\right)\left(\sum_{n=0}^\infty (ab_{d+1})^n\right)\\ &= \sum_{n=0}^\infty \left(\sum_{l=0}^n \left(\sum_{c_1+ \dots +c_d=l}b_1^{c_1} \dots b_d^{c_d}\right)b_{d+1}^{n-l}\right)a^n\\ &=\sum_{n=0}^\infty \left(\sum_{c_1+\dots + c_{d+1}=n}b_1^{c_1} \dots b_{d+1}^{c_{d+1}}\right) a^n \end{aligned}$$

  • $\begingroup$ Thank you for your answer, I see how this can be done with a product of two series but I fail to reproduce this with three or higher series, any tips? $\endgroup$
    – Chris
    Commented Nov 22, 2020 at 14:03
  • 1
    $\begingroup$ Answer updated. $\endgroup$ Commented Nov 22, 2020 at 14:16

We introduce the indicator function notation defined by

$$ \mathbf{1}_{\{ \cdots \}} = \begin{cases} 1, & \text{if $\cdots$ holds}, \\ 0, & \text{if $\cdots$ does not hold}. \end{cases} $$

Then the sum is can be rewritten as

\begin{align*} \sum_{n=0}^{\infty}a^n \sum_{\substack{c_1,\ldots,c_d\geq0\\c_1+\cdots+c_d=n}} b_1^{c_1} \cdots b_d^{c_d} &= \sum_{n=0}^{\infty} \sum_{\substack{c_1,\ldots,c_d\geq0\\c_1+\cdots+c_d=n}} (a b_1)^{c_1} \cdots (a b_d)^{c_d} \\ &= \sum_{n=0}^{\infty} \sum_{c_1,\ldots,c_d\geq0} \mathbf{1}_{\{ c_1+\cdots+c_d=n \}}(a b_1)^{c_1} \cdots (a b_d)^{c_d}. \end{align*}

Since the two sigma-notations in the last line are independent of each other, we can freely interchange the order[1] to get

\begin{align*} \sum_{c_1,\ldots,c_d\geq0} \Biggl( \sum_{n=0}^{\infty} \mathbf{1}_{\{ c_1+\cdots+c_d=n \}} \Biggr) (a b_1)^{c_1} \cdots (a b_d)^{c_d}. \end{align*}

However, for each given $c_0, \ldots, c_d \geq 0$, there is exactly one $n \geq 0$ such that $c_1+\cdots+c_d=n$ holds, and so, the inner sum is always $1$. So the sum simplifies to

$$ \sum_{c_1,\ldots,c_d\geq0} (a b_1)^{c_1} \cdots (a b_d)^{c_d} = \Biggl( \sum_{c_1 \geq 0} (a b_1)^{c_1} \Biggr) \cdots \Biggl( \sum_{c_d \geq 0} (a b_d)^{c_d} \Biggr) = \frac{1}{1 - ab_1} \cdots \frac{1}{1 - ab_d}. $$

Remark. The same conclusion can be derived without the use of indicator function. However, my personal impression is that this method provides great flexibility when working with multiple sums.

[1] Switching the order of summation is not always possible. Fubini's theorem provides a sufficient condition for this. In OP's case, this is possible if either $\left| ab_i \right| < 1$ for all $i = 1, \ldots, d$ or $ab_i \geq 0$ for all $i = 1, \ldots, d$. Alternatively, if we regard the sum as a formal power series, then no such assumption is required.


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