Show that $\lim_{x\rightarrow 1}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x}=\ln2$. Prove $$\lim_{x\rightarrow 1}\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}=\ln2.$$

Of course  $$\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}}=\ln2,$$
but we can not use the Proposition :
If a sequence of functions that are continuous on a set converges uniformly on that set ,then the limit function is continuous on the set.
because for $\forall\delta >0$,we have $\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}} $converges on$\left(1-\delta ,1+\delta  \right), $but nonuniformly .

In fact ,for every ${x}^{'},{x}^{"}\in \left(1,1+\delta  \right)，$in other words $\left| {x}^{'}-{x}^{"}\right|<\delta  $,fixing  the point $ {x}^{'}, $let ${x}^{"}\rightarrow {1}^{+},$we can get paradox:$$1\geq |\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{{x}^{'}}}-(+\infty)|=+\infty>1.$$
So $\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}$ converge nonuniformly on $\left(1-\delta  ,1+\delta  \right)$.

My question is how can we get $\lim_{x\rightarrow 1}\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}=\ln2.$ 
 A: Let
$$f_N(x) = \sum_{n=1}^N \frac{(-1)^{n-1}}{n^x},\quad R_N(x) = \sum_{n=N+1}^\infty \frac{(-1)^{n-1}}{n^x}.$$
It is clear that for every $N$ you have $\lim_{x\to 1} f_N(x) = f_N(1)$. It remains to find a bound on
$$\lvert R_N(x)\rvert.$$
The fact that the series is alternating should help with that. (And that shows that actually the convergence of the series is uniform on $(1-\delta,1+\delta)$ for $0 < \delta < 1$.)

 Namely, for $x > 0$, we have $\frac{1}{n^x} > \frac{1}{(n+1)^x}$, and thus by Leibniz' criterion for alternating series, it follows that $$\lvert R_N(x)\rvert = \left\lvert \sum_{k=0}^\infty \frac{(-1)^k}{(N+1+k)^x}\right\rvert \leqslant \frac{1}{(N+1)^x}.$$ Since $x \mapsto \frac{1}{(N+1)^x}$ is monotonically decreasing on $(0,+\infty)$ for every $N\in\mathbb{N}\setminus \{0\}$, we thus have $$\lvert R_N(x)\rvert \leqslant \frac{1}{(N+1)^x} \leqslant \frac{1}{(N+1)^{1-\delta}}$$ for all $N$ and $x \geqslant 1-\delta$. It follows that the convergence of $f_N$ is uniform on every $[1-\delta,+\infty)$ for $0 < \delta < 1$. Thus the proposition cited in the question can be used.

A: You may observe that
$$
\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}=\frac{1}{\Gamma(x)}\int_0^\infty\frac{u^{x-1}}{e^u+1} \mathrm{d}u, \quad x>1.
$$
Letting $ x \rightarrow 1^+$ gives
$$
\lim_{x\rightarrow 1^+}\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{{n}^{x}}=\int_0^\infty\frac{1}{e^u+1} \mathrm{d}u=\ln 2 .
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{}$
\begin{align}&\color{#66f}{\large\lim_{x\ \to\ 1}\sum_{n = 1}^{\infty}%
{\pars{-1}^{n - 1} \over n^{x}}}
=-\lim_{x\ \to\ 1}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{x}}
=-\lim_{x\ \to\ 1}{\rm Li}_{x}\pars{-1}=-\,{\rm Li}_{1}\pars{-1}
\\[3mm]&=-\braces{-\ln\pars{1 - \bracks{-1}}}
=\color{#66f}{\Large\ln\pars{2}} \approx {\tt 0.6931}
\end{align}
