I'm looking at a complex function defined as $$f(z) = \sum\limits_{r=1}^\infty (-1)^{r+1}\sin\left(\frac{pz}{r}\right)$$ and am looking to find its Taylor expansion about $z=0$. Presumably $p$ is some real number.
My instinct is to Taylor expand $\sin$ about $z=0$, then combine the sums. The only issue is that I have no idea how to combine sums nicely. Is this a decent approach, or is there a better way?
One approach is to find the derivatives of $f$ at zero. Clearly $f^{(k)}(0) = 0$ for all even $k$. If $k \equiv 1 \pmod 4$, then we have $$ f^{(k)}(0) = \sum_{r=1}^\infty (-1)^{r+1} (p/r)^k = p^k\sum_{r=1}^\infty \frac{(-1)^{r+1}}{r^k}. $$ If $k \equiv 3 \pmod 4$, then we have $$ f^{(k)}(0) = -\sum_{r=1}^\infty (-1)^{r+1} (p/r)^k = -p^k\sum_{r=1}^\infty \frac{(-1)^{r+1}}{r^k}. $$ With that, we can write $f(z) = \sum_{n=0}^\infty (-1)^n \eta(2n+1) p^{2n+1}z^{2n + 1}$, where $$ \eta(k) = \sum_{r=1}^\infty \frac{(-1)^{r+1}}{r^k}. $$ Note that this $\eta$ is known as the Dirichlet eta function.