Showing the quintessential logarithm property using the Maclaurin series of $\log$

Question

For $-1\le x<1$, we have $$\log(1-x) = -\sum_{k=1}^{\infty} \frac{x^{k}}{k}\\$$ Taking $a,b$ with $|a|,|b|<1$ and $(1-a)(1-b)\le2$, on the function side clearly we have $$\log(1-a)+\log(1-b) = \log((1-a)(1-b)) = \log(1-(a+b-ab))$$ On the series side, dispensing with the negatives this becomes $$\sum_{k=1}^{\infty} \frac{a^{k}+b^k}{k}=\sum_{k=1}^{\infty} \frac{(a+b-ab)^{k}}{k}$$ This feels decidedly less-obvious: in order for us to have equality, all the mixed terms should cancel when we expand the sum. I have tried using the multinomial theorem to write the RHS as a multi-sum and tried several things in that vein to no end. For instance, we would write the RHS as: $$\sum_{k=1}^{\infty} \frac{1}{k}\cdot\left(\sum_{\substack{k_1+k_2+k_3=k\\ k_i\ge 0}} \binom{k}{k_1,k_2,k_3} a^{k_1}b^{k_2}(-ab)^{k_3}\right)$$ $$=\sum_{k=1}^{\infty} \frac{1}{k}\cdot\left(\sum_{\substack{k_1+k_2+k_3=k\\ k_i\ge 0}} \binom{k}{k_1,k_2,k_3} a^{k_1+k_3}b^{k_2+k_3}(-1)^{k_3}\right)$$ $$=\sum_{k=1}^{\infty} \frac{1}{k}\cdot\left(a^k+b^k+(-ab)^k+\sum_{\substack{k_1+k_2+k_3=k\\ k_i\ge 1}} \binom{k}{k_1,k_2,k_3} a^{k_1+k_3}b^{k_2+k_3}(-1)^{k_3}\right)$$ Then it would suffice to show $$-\sum_{k=1}^{\infty} \frac{(-ab)^k}{k}=\sum_{k=1}^{\infty} \frac{1}{k}\cdot\left(\sum_{\substack{k_1+k_2+k_3=k\\ k_i\ge 1}} \binom{k}{k_1,k_2,k_3} a^{k-k_2}b^{k-k_1}(-1)^{k_3}\right),$$and factoring out $(ab)^k$ and simplifying leaves $$\log(2)=\sum_{k=1}^{\infty} \frac{1}{k}\cdot\left(\sum_{\substack{k_1+k_2+k_3=k\\ k_i\ge 1}} \binom{k}{k_1,k_2,k_3} a^{-k_2}b^{-k_1}(-1)^{k_3}\right),$$ but I cannot make progress here. Of course, the result is obvious if one uses the closed-form of the logarithm, but the point of this question is to avoid doing that.

Answer 1

One method is to use Lucas sequences. It is defined as $\,V_n := a^n+b^n.\,$ Define the quantities $\,P:=a+b,$ $Q:=ab.\,$ Then the recursion is $\,V_{n+2} = P V_{n+1} - QV_n.\,$ This implies that $\,V_n\,$ is a polynomial in $\,P,Q.\,$ First, define the sequence:

$$ S_{n,m} := {n+m-1 \choose m-1}\frac1{m} P^n(-Q)^m. \tag1 $$

Then, use a formula from OEIS A034807 to get

$$ A_k := \frac{a^k+b^k}k = \frac{V_k}k = \frac{P^k}k + \sum_{j=1}^{k/2}S_{k-2j,j}. \tag2 $$

Use the binomial theorem on $\,(P-Q)^k\,$ to get

$$ B_k := \frac{(a+b-ab)^{k}}{k} = \frac{(P-Q)^k}{k} = \frac{P^k}k + \sum_{j=1}^k S_{k-j,j}. \tag3 $$

Notice that \begin{align*} \sum_{k=1}^\infty A_{k} &= \sum_{k=1}^\infty \frac{P^k}k + \sum_{k=1}^\infty \sum_{j=1}^{k/2}S_{k-2j,j} \\ &= -\log(1-P) + \sum_{n=0}^\infty\sum_{m=1}^\infty S_{n,m} \tag4 \end{align*}

where $\,n = k-2j,\, m = j.\,$ Similarly, \begin{align*} \sum_{k=1}^\infty B_{k} &= \sum_{k=1}^\infty \frac{P^k}k + \sum_{k=1}^\infty \sum_{j=1}^k S_{k-j,j} \\ &= -\log(1-P) + \sum_{n=0}^\infty\sum_{m=1}^\infty S_{n,m} \tag5 \end{align*}

where $\,n = k-j, m = j.\,$ Combine the two equations to get

$$ \sum_{k=1}^\infty A_{k} = \sum_{k=1}^\infty B_{k}. $$