TITLE: Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$
QUESTION [5 upvotes]: If $\ln(1-x+x^2)  = a_1x+a_2x^2 + \cdots \text{ then } a_3+a_6+a_9+a_{12} + \cdots = $ ?
My approach is to write $1-x+x^2 = \frac{1+x^3}{1+x}$ then expanding the respective logarithms,I got a series (of coefficient) which is nothing but $\frac{2}{3}\ln 2$.But this approach took some time for me (I can't solved it during the test but after the test I solved it)... Any other quick method?

REPLY [4 votes]: Qiaochu Yuan gives an excellent exposition of the general method. Sometimes in specific cases you can get the answer via some hands-on calculations as well, and expanding the logs as you mentioned can actually be done pretty quickly: using that $1 - x + x^2 = {1 + x^3 \over 1 + x}$ one gets that $\ln(1 - x + x^2) = \ln(1 + x^3) - \ln(1 + x)$. You now can use the fact that the sum of $a_3 + a_6 + ...$ is obtained by subtracting the corresponding sums for $\ln(1 + x^3)$ and $\ln(1 + x)$. Since the power series for $\ln(1 + x^3)$ only contains powers of $x^3$, the contribution to $a_3 + a_6 + ...$ coming from that term is what you get from plugging in $x = 1$, namely $\ln(2)$.
The power series of $\ln(1 + x)$ may be written as $-\sum_{n > 0} {(-x)^n \over n}$.
Taking every third term gives $-\sum_{n > 0}  {(-x)^{3n} \over 3n} = -{1 \over 3}\sum_{n > 0} {(-x^3)^n \over n}$. Note this series is again the series of $\ln(1 + x)$, but applied to $x^3$ in place of $x$. Plugging in $x = 1$ gives ${1 \over 3}\ln(2)$.
So the answer you want is $\ln(2) - {1 \over 3}\ln(2) = {2 \over 3}\ln(2)$.