Prove that $\lim\limits_{x\to 0^+}\sum\limits_{n=1}^\infty\frac{(-1)^n}{n^x}=-\frac12$ 
Prove that $$\lim\limits_{x\to 0^+}\sum\limits_{n=1}^\infty\frac{(-1)^n}{n^x}=-\frac12$$

Does the limit even exist? Because when I develop the sum and consider the limit $x\to 0^+$, I get the sum $1-1+1-1+1-\dots$
 A: We can write this sum in terms of the Dirichlet eta function or the Riemann zeta function:
$$
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^n}{n^z}
&=-\eta(z)\\
&=-\left(1-2^{1-z}\right)\zeta(z)\tag1
\end{align}
$$
In this answer it is shown that $\zeta(0)=-\frac12$. Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{z\to0}\sum_{n=1}^\infty\frac{(-1)^n}{n^z}=-\frac12}\tag2
$$

Computing $\boldsymbol{\eta'(0)}$
Using the formula $\eta(z)\Gamma(z)=\int_0^\infty\frac{\,t^{z-1}}{e^t+1}\,\mathrm{d}t$, we get
$$
\begin{align}
\eta(z)\Gamma(z+1)
&=\int_0^\infty\frac{z\,t^{z-1}}{e^t+1}\,\mathrm{d}t\\
&=\int_0^\infty\frac1{e^t+1}\,\mathrm{d}t^z\\
&=\int_0^\infty\frac{t^z\,e^t}{\left(e^t+1\right)^2}\,\mathrm{d}t\tag3
\end{align}
$$
As shown in this answer,
$$
\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=-\gamma\tag4
$$
Using Gautschi's Inequality, $\lim\limits_{n\to\infty}\frac{\Gamma(n+1)}{\Gamma\left(n+\frac12\right)}
\frac{\Gamma(n+1)}{\Gamma\left(n+\frac32\right)}=1$; therefore,
$$
\begin{align}
\prod\limits_{k=1}^\infty\frac{2k}{2k-1}\frac{2k}{2k+1}
&=\lim_{n\to\infty}\prod\limits_{k=1}^n\frac{k}{k-\frac12}\frac{k}{k+\frac12}\\
&=\lim_{n\to\infty}\frac{\Gamma(n+1)\,\color{#090}{\Gamma\left(\frac12\right)}}{\Gamma\left(n+\frac12\right)}
\frac{\Gamma(n+1)\,\color{#090}{\Gamma\left(\frac32\right)}}{\Gamma\left(n+\frac32\right)}\\[6pt]
&=\color{#090}{\frac\pi2}\tag5
\end{align}
$$
Taking the derivative of $(3)$ and evaluating at $z=0$, we have
$$
\begin{align}
\eta'(0)-\frac12\gamma
&=\int_0^\infty\log(t)\frac{e^{-t}}{(1+e^{-t})^2}\,\mathrm{d}t\tag{6a}\\
\eta'(0)
&=-\int_0^\infty\log(t)\,\mathrm{d}\left(\frac{e^{-t}}{1+e^{-t}}-\frac{e^{-t}}2\right)\tag{6b}\\
&=-\frac12\int_0^\infty\log(t)\,\mathrm{d}\frac{e^{-t}(1-e^{-t})}{1+e^{-t}}\tag{6c}\\
&=\frac12\int_0^\infty\frac{e^{-t}(1-e^{-t})}{t\left(1+e^{-t}\right)}\,\mathrm{d}t\tag{6d}\\
&=\frac12\sum_{k=1}^\infty(-1)^{k-1}\int_0^\infty\frac{e^{-kt}-e^{-(k+1)t}}t\,\mathrm{d}t\tag{6e}\\
&=\frac12\sum_{k=1}^\infty(-1)^{k-1}\log\left(\frac{k+1}k\right)\tag{6f}\\
&=\frac12\sum_{k=1}^\infty\log\left(\frac{2k}{2k-1}\frac{2k}{2k+1}\right)\tag{6g}\\
&=\frac12\log\left(\frac\pi2\right)\tag{6h}
\end{align}
$$
Explanation:
$\text{(6a)}$: taking the derivative of $(3)$ and evaluating at $z=0$
$\text{(6b)}$: prepare to integrate by parts and subtract half of $(4)$ from both sides
$\text{(6c)}$: combine fractions
$\text{(6d)}$: integrate by parts
$\text{(6e)}$: write $\frac{e^{-t}}{1+e^{-t}}$ as a series in $e^{-t}$
$\text{(6f)}$: apply Frullani's Integral
$\text{(6g)}$: combine adjacent positive and negative terms
$\text{(6h)}$: apply $(5)$
Equation $(6)$ justifies Claude Leibovici's comment on the question that for $z$ slightly above $0$,
$$
\sum_{n=1}^\infty\frac{(-1)^n}{n^z}\sim-\frac12-\frac z2\,\log\left(\frac\pi2\right)\tag7
$$
A: I'll look at $\sum_{n=1}^{\infty}(-1)^{n+1}/n^x$ instead. (I like starting with a positive summand!) We'll then want to show the limit is $1/2.$
This sum equals
$$\tag 1 \sum_{n=1}^{\infty}\left (\frac{1}{(2n-1)^x} - \frac{1}{(2n)^x} \right ) = \sum_{n=1}^{\infty}\frac{(2n)^x -(2n-1)^x }{(2n-1)^x(2n)^x}.$$
Now $(2n)^x -(2n-1)^x = xc^{x-1}$ by the MVT. Here $2n-1<c<2n.$ For $0<x<1$ we'll have $c^{x-1}> (2n)^{x-1}.$ It follows that $(1)$ is bounded below by
$$\tag 2\sum_{n=1}^{\infty}\frac{x}{(2n)^{1-x}(2n)^{x}(2n)^{x}} = x\sum_{n=1}^{\infty}\frac{1}{(2n)^{1+x}}  > x \int_1^\infty (2t)^{-1-x} \, dt = x\cdot\frac{1}{x2^{1+x}}.$$
We get a bound from above using $c^{x-1}< (2n-1)^{x-1}.$ This shows $(1)$ is less than
$$\tag 3 x\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{1+x}} < x(1 + \int_1^\infty (2t-1)^{-1-x} \, dt) = x\left(1+\frac{1}{2x}\right).$$
As $x\to 0^+,$ the limits of the terms on the right of $(2)$ and $(3)$ are $1/2,$ and we're done.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{x \to 0^{\large +}}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^x} =
-\,{1 \over 2}:\ {\large ?}}$

\begin{align}
\lim_{x \to 0^{\large +}}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^x} & =
\lim_{x \to 0^{\large +}}\mrm{Li}_{x}\pars{-1} 
\end{align}

$\ds{\mrm{Li}_{s}}$ is the Order-$\ds{s}$
  Polylogarithm.

By using its
integral representation: 
\begin{align}
\lim_{x \to 0^{\large +}}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^x} & =
-\lim_{x \to 0^{\large +}}\bracks{%
{1 \over \Gamma\pars{x}}\int_{0}^{\infty}{t^{x - 1} \over \expo{t} + 1}\,\dd t} \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
-\lim_{x \to 0^{\large +}}\braces{%
{1 \over \Gamma\pars{x}}\bracks{%
{1 \over x}\int_{0}^{\infty}{t^{x}\expo{t} \over \pars{\expo{t} + 1}^{2}}
\,\dd t}}
\\[5mm] & =
-\lim_{x \to 0^{\large +}}\bracks{%
{1 \over \Gamma\pars{x + 1}}
\int_{0}^{\infty}{t^{x}\expo{t} \over \pars{\expo{t} + 1}^{2}}\,\dd t} =
-\int_{0}^{\infty}{\expo{t} \over \pars{\expo{t} + 1}^{2}}\,\dd t
\\[5mm] & =
\left.{1 \over \expo{t} + 1}\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} =
\bbx{-\,{1 \over 2}}
\end{align}
