How to prove the limit related to the following infinite series?

Question

I would appreciate it if you can tell me how to prove this limit $$\lim_{x\to 0^{+}}\sum_{n=2}^{\infty}\frac{(-1)^{n}}{(\log n)^{x}}=\frac{1}{2}$$

Answer 1

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 $\int_0^\infty\eta'(y)~dy=\eta(\infty)-\eta(0^+)=1/2$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}