Does the following infinite series:
$$G(s):=\displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n+1}}{n^s + \frac{1}{n^s}}$$
converge for all $s \in \mathbb{C}$ (especially in the critical strip; since when $\Re(s) \rightarrow \infty$, then $G(s) \rightarrow \frac12$ ) ?
Note that $G(s)=G(-s)$. EDIT Based on good feedback in the comments: the series clearly diverges for $s=0$. A possible analytic continuation of $G(s)$ at that point, could be to use Grandi's series, that has a Cesàro sum of $\frac12$, hence $G(0)$ might be 'assigned' a value of $\frac14$.
I don't know about on the imaginary axis, but $G(s)$ converges for all $s$ with nonzero real part. Since $G(-s)=G(s)$, it suffices to consider $\Re s>0$. Choose an integer $k\ge(2\Re s)^{-1}$. Then \begin{align*} G(s) &= \sum_{n=1}^\infty (-1)^{n+1} n^{-s}(1+n^{-2s})^{-1}\\ &= \sum_{n=1}^\infty (-1)^{n+1} \bigg( \sum_{j=1}^k (-1)^{j-1} n^{-(2j-1)s} + \bigg( n^{-s}(1+n^{-2s})^{-1} - \sum_{j=1}^k (-1)^{j-1} n^{-(2j-1)s} \bigg) \bigg) \\ &= \sum_{j=1}^k (-1)^j \sum_{n=1}^\infty (-1)^n n^{-(2j-1)s} - \sum_{n=1}^\infty (-1)^n \bigg( n^{-s}(1+n^{-2s})^{-1} - \sum_{j=1}^k (-1)^{j-1} n^{-(2j-1)s} \bigg), \end{align*} assuming that these $k+1$ infinite series all converge. But in the first sum, each of the $k$ inner sums converges for real $s>0$ by the alternating series test, and hence converges for all $s$ with $\Re s>0$ by the general theory of Dirichlet series. As for the second sum, the $n$th term is $O(n^{-(2k+1)s})$, and so that series converges absolutely at $s$ by our choice of $k$.