 Solution

Near zero, expand $e^z-1=z(1+z/2+z^2/6+\cdots )$ and invert the parenthesized series. The first three nonzero Laurent series terms areThe zero quadratic coefficient also follows from $f(-z)=-1-f(z)$, which makes $f(z)+1/2$ odd. The nearest other singularities are at $\pm 2\pi i$.

All poles of $f$ occur at $2\pi ir$, $r\in \mathbb{Z}$, and have residue one, since the derivative of $e^z-1$ equals one at those zeros. AlsoThus $f_1$ subtracts the complete principal part at zero and at every pole with $1\le |r|\le n$. These are all the poles in the stated open disk; each singularity of $f_2=f-f_1$ there is removable.

For $|z|>2n\pi$, expanding each rational summand in inverse powers givesThis contributes no constant or positive-power term. To find the first Taylor coefficients of the analytic $f_2$, expand $f_1$ near zero as $z^{-1}+z{\sum }_{r=1}^n(2{\pi }^2{r}^2{)}^{-1}+O(z^3)$ and subtract it from the already calculated local series of $f$. Combining its Taylor series with the exterior series above yields the three requested annular coefficients:Both series converge on $2n\pi <|z|<2(n+1)\pi$, so this combination is a valid Laurent expansion there.

For the Basel sum from exponential poles, let $R_n=(2n+1)\pi$. The Laurent coefficient formula givesThe function $f$ is bounded on these circles by one constant independent of $n$. Indeed, if $\mathrm{Re}z\ge 1$, use $|e^z-1|\ge e-1$; if $\mathrm{Re}z\le -1$, use $|e^z-1|\ge 1-e^{-1}$. In the remaining strip $|\mathrm{Re}z|\le 1$, the imaginary part has magnitude $\sqrt{R_n^2-(\mathrm{Re}z{)}^2}$, within $1/R_n$ of the odd multiple $R_n$ of $\pi$. Its cosine is negative, so $|e^z-1{|}^2=e^{2\mathrm{Re}z}+1-2e^{\mathrm{Re}z}\cos (\mathrm{Im}z)\ge 1$.

The contour length is $2\pi R_n$, hence $|a_1|\le C/R_n\to 0$. Taking the limit in the coefficient formula proves