Is there an elementary way to analytically continue
$$f(s)=\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)^s}$$
to the entire complex plane? It is not hard to see (by grouping terms in pairs and using the mean value theorem) that it converges for $\Re(z)>0$.
I realize there is a vast literature on Dirichlet series that almost certainly answers this with some general theorem. However, this sum appears as a qualifying exam question (1.43), so presumably there is a more direct method. For example, if the terms did not alternate, we could just take $(1-2^{-s})\zeta(s)$ and use the fact $\zeta(s)$ has a continuation to the entire plane.
Assuming $\text{Re}(s)>0$ we have:
$$\begin{eqnarray*} f(s)=\sum_{n\geq 1}\frac{(-1)^n}{(2n+1)^s}&=&\sum_{n\geq 1}\frac{(-1)^n}{\Gamma(s)}\int_{0}^{1}(-\log x)^s x^{2n}\,dx\\&=&-\frac{1}{\Gamma(s)}\int_{0}^{1}(-\log x)^s \frac{x^2}{1+x^2}\,dx\\&=&-\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{t^s}{e^t(1+e^{2t})}\,dt\end{eqnarray*} $$ and we may use integration by parts to find an analytic continuation over $\text{Re}(s)>-1$.
After that, we just have to repeat the process.