Abscissa of convergence of powers of alternating zeta function Consider the Dirichlet series given by alternating zeta function: 
$$\eta(s) := \sum_1^{\infty} (-1)^{n+1} n^{-s}.$$
On page 19 of 'Multiplicative Number Theory' by Montgomery and Vaughan, it is stated that the abscissa of convergence of the Dirichlet series
$$[\eta(s)]^2$$
is at least $\frac{1}{4}$. 
(a) Is there a simple proof of this fact? 
(b) This statement suggests that for any positive integer $k \geq 2$, the abscissa of convergence of $$[\eta(s)]^k$$ is $\alpha_k$, the order of magnitude of the remainder term in the $k$-divisor problem (since it is known that $\alpha_2 \geq \frac{1}{4}$). Does anyone have a proof or disproof?
Thank you.
 A: The Dirichlet series for $\eta(s)^2$ equivalently (Dirichlet divisor problem) the Dirichlet series for $\zeta(s)^2 - (2\gamma-1)\zeta(s)+\zeta'(s)$ converge for $\Re(s) > \theta$ where it is easily shown $\theta \le 1/2$ and it is a theorem of Hardy that $\theta \ge 1/4$. 
There is a whole chapter in Titchmarsh about the approach based on the mean growth of $\zeta(s)$ and $|\zeta(s)|^k$ on vertical lines.
The current knowledge is $\theta\in [1/4, 131/416] $.

In Titchmarsh the proof of $\theta \ge 1/4$ is that for $\sigma > \theta$ (Parseval theorem for the Mellin/Fourier transform $\zeta(s)^2 = s \int_1^\infty \Delta(x) x^{-s-1}dx$) then $$\frac1{2\pi}\int_{-\infty}^\infty \frac{|\zeta(\sigma+it)|^4}{|\sigma+it|^2}dt=\int_0^\infty \Delta(x)^2 x^{-2 \sigma-1}dx <\infty$$
$\int_{-\infty}^\infty \frac{|\zeta(1/4+it)|^4}{|1/4+it|^2}dt< \infty$ can be disproved from the functional equation and $T \zeta(3/4) \sim \int_T^{2T} |\zeta(3/4+it)|^2 dt \le (\int_T^{2T} |\zeta(3/4+it)|^4 dt)^{1/2} (\int_T^{2T} dt)^{1/2}$
