convergence of $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{s}} \cos(\ln(n) t)$ for $s>0$ I'm trying to figure out whether this series converges for $s>0$ for any given value of t. I've found a few things on here for given values of $t$, notably $t=1$, but I'm struggling to find a proof for all t. So far I've tried using the comparison test to reduce the sequence to one thats more manageable, but because the sign alternates I'm unable to do this. Does anyone have an elementary way to approach this?
 A: Notice that if $a_n = (-1)^{n+1}$, then your series is the real part of the Dirichlet series
$$\zeta \big( (a_n)_{n \ge 1}, z \big) = \sum _{n \ge 1} \frac {a_n} {n^z}$$
for $z = s + \Bbb i t$.
Since $(a_n)_{n \ge 1}$ is bounded, it is known that $\zeta \big( (a_n)_{n \ge 1}, z \big)$ is absolutely convergent for $s>1$.
If $A(N) = \sum _{n=1} ^N a_n = \begin{cases} 0, & 2 \mid N \\ 1, & 2 \nmid N \end{cases}$ (notice that $\big( A(N) \big) _{N \ge 1}$ diverges) then the abscisssa of convergence of $\zeta \big( (a_n)_{n \ge 1}, z \big)$ is
$$\limsup _{N \to \infty} \frac {\ln |A(N) - 0|} {\ln N} = \limsup _{N \to \infty} \frac {\ln 1} {\ln N} = 0 ,$$
whic means that for $s \in (0,1)$ the series $\zeta \big( (a_n)_{n \ge 1}, z \big)$ converges but not absolutely.
Of course, all the conclusions obtained for $\zeta \big( (a_n)_{n \ge 1}, z \big)$ remain true for just its real part, which is exactly your series.

More simply, though, you could say that your function is the real part of Dirichlet's $\eta$ function, and is therefore known to converge only for $s>0$.
