How to prove the limit related to the following infinite series? I would appreciate it if you can tell me how to prove this limit
$$\lim_{x\to 0^{+}}\sum_{n=2}^{\infty}\frac{(-1)^{n}}{(\log n)^{x}}=\frac{1}{2}$$
 A: Using $\int_0^\infty y^{a-1}e^{-by}~dy=\Gamma(a)/b^a$ for $a,b>0$, and denoting $\eta(y)$${}=\sum_{n=1}^\infty(-1)^{n-1}n^{-y}$, we get $$\sum_{n=2}^\infty\frac{(-1)^n}{(\ln n)^x}=\frac{1}{\Gamma(x)}\int_0^\infty y^{x-1}\big(1-\eta(y)\big)~dy=\frac{1}{\Gamma(1+x)}\int_0^\infty y^x\eta'(y)~dy$$ (the last equality comes from integration by parts). Since, if not considered known, $$\eta(y)=\frac{1}{\Gamma(y)}\int_0^\infty\frac{z^{y-1}~dz}{e^z+1}=\frac{1}{\Gamma(1+y)}\int_0^\infty\frac{z^y e^z~dz}{(e^z+1)^2}\ \color{blue}{\underset{y\to0^+}{\longrightarrow}}\ \int_0^\infty\frac{e^z~dz}{(e^z+1)^2}=\frac{1}{2}$$ along the same lines, the required limit is $\int_0^\infty\eta'(y)~dy=\eta(\infty)-\eta(0^+)=1/2$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\lim_{x \to 0^{+}}\,\,\sum_{n = 2}^{\infty}
{\pars{-1}^{n} \over \ln^{x}\pars{n}}} =
\lim_{x \to 0^{+}}\,\,\sum_{n = 2}^{\infty}
\pars{-1}^{n}\bracks{{1 \over \Gamma\pars{x}}
\int_{0}^{\infty}t^{x - 1}\expo{-\ln\pars{n}t}\,\dd t}
\\[5mm] = &\
\lim_{x \to 0^{+}}\,\,{x \over \Gamma\pars{x + 1}}\int_{0}^{\infty}t^{x - 1}
\sum_{n = 2}^{\infty}{\pars{-1}^{n} \over n^{t}}\,\dd t
\\[5mm] = &\
\lim_{x \to 0^{+}}\,\,x\int_{0}^{\infty}t^{x - 1}
\bracks{1 - \pars{1 - 2^{1 - t}}\zeta\pars{t}}\,\dd t
\\[5mm] \stackrel{\mrm{IBP}}{=} &\
-\lim_{x \to 0^{+}}\,\,\int_{0}^{\infty}t^{x}
\totald{\bracks{1 - \pars{1 - 2^{1 - t}}\zeta\pars{t}}}{t}\,\dd t
\\[5mm] = &\
-\int_{0}^{\infty}
\totald{\bracks{1 - \pars{1 - 2^{1 - t}}\zeta\pars{t}}}{t}\,\dd t
\\[5mm] = &\
-\ \underbrace{\lim_{t \to \infty}
\bracks{1 - \pars{1 - 2^{1 - t}}\zeta\pars{t}}}_{\ds{=\ 0}}\ +\
\underbrace{\lim_{t \to 0}
\bracks{1 - \pars{1 - 2^{1 - t}}\zeta\pars{t}}}
_{\ds{=\ \color{red}{1 \over 2}}}
\\[5mm] = &\ \bbx{1 \over 2} \\ &
\end{align}
