What is the value of the limit $\lim_{a\searrow 0}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{a}}$? Clearly the series
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{a}}
$$
converges (conditionally), as an alternating series of as absolutely decreasing sequence, for all $a>0$. 
The question is: What is the value of 
$$
\lim_{a\searrow 0}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{a}}.
$$
First attempt
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{a}}=\frac{a}{2}\sum_{n=1}^\infty \frac{1}{n^{a+1}}+{\mathcal O}(a).
$$
 A: The series is, for $\operatorname{Re} a > 0$, the Dirichlet $\eta$ function,
$$\eta(a) = \left(1-2^{1-a}\right)\zeta(a).$$
Since $\eta$ is an entire function, by continuity, the limit is
$$\eta(0) = -\zeta(0) = \frac{1}{2}.$$
A: The answer of D. Fischer is shorter and elegant. Let me nevertheless provide my own elementary answer:
First note that
$$
\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^a}-\frac{1}{(2n)^a}\right)=\sum_{n=1}^\infty \left(\frac{(2n)^a-(2n-1)^a}{(2n-1)^a(2n)^a}\right)=\sum_{n=1}^\infty \frac{(1+\frac{1}{2n-1})^a-1}{(2n)^a} \\
=\sum_{n=1}^\infty\frac{\frac{a}{2n-1}+\frac{a(a-1)}{2(2n-1)^2}+\cdots}{(2n)^a}=\cdots=\frac{a}{2}\sum_{n=1}^\infty
\frac{1}{n^{a+1}}+{\mathcal O}(a).
$$
Next obrserve that, there exists $\xi,\zeta\in (0,1)$, such that
$$
\frac{1}{n^{a+1}}-\int_{n}^{n+1}\frac{dx}{x^{a+1}}=\frac{1}{n^{a+1}}-\frac{1}{(n+\xi)^{a+1}}=
\frac{\xi(a+1)}{(n+\zeta)^{a+2}},
$$
and hence
$$
0<\frac{1}{n^{a+1}}-\int_{n}^{n+1}\frac{dx}{x^{a+1}}< 
\frac{a+1}{n^{2}},
$$
which implies that
$$
\frac{a}{2}\sum_{n=1}^\infty\frac{1}{n^{a+1}}=\frac{a}{2}\int_1^\infty\frac{dx}{x^{a+1}}+{\mathcal O}(a).
$$
But
$$
\frac{a}{2}\int_1^\infty\frac{dx}{x^{a+1}}=\lim_{M\to\infty}\frac{a}{2}\left(\frac{1}{a}-\frac{1}{aM^a}\right)=\frac{1}{2},
$$
which implies that
$$
\lim_{a\searrow 0}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{a}}=\frac{1}{2}.
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{n\in{\mathbb N}\,,\ N\ \geq\ 1}$:

  
*
*
  \begin{align}\boxed{\ds{\sum_{n\ =\ 1}^{2N}{\pars{-1}^{n - 1} \over n^{a}}}}
&=\sum_{n\ =\ 1}^{N}{1 \over \pars{2n - 1}^{a}}
-\sum_{n\ =\ 1}^{N}{1 \over \pars{2n}^{a}}
\\[5mm]&=\sum_{n\ =\ 1}^{N}{1 \over n^{a}}
-\sum_{n\ =\ 1}^{N}{1 \over \pars{2n}^{a}}
-\sum_{n\ =\ 1}^{N}{1 \over \pars{2n}^{a}}
=\boxed{\ds{\pars{1 - 2^{1 - a}}\sum_{n\ =\ 1}^{N}{1 \over n^{a}}}}
\end{align}
  
  
*Similarly,
  \begin{align}\boxed{\ds{\sum_{n\ =\ 1}^{2N - 1}{\pars{-1}^{n - 1} \over n^{a}}}}
&=\sum_{n\ =\ 1}^{N}{1 \over \pars{2n - 1}^{a}}
-\sum_{n\ =\ 1}^{N - 1}{1 \over \pars{2n}^{a}}
\\[5mm]&=\sum_{n\ =\ 1}^{N}{1 \over n^{a}}
-\sum_{n\ =\ 1}^{N}{1 \over \pars{2n}^{a}}
-\sum_{n\ =\ 1}^{N}{1 \over \pars{2n}^{a}} + {1 \over \pars{2N}^{a}}
\\[5mm]&=\boxed{\ds{\pars{1 - 2^{1 - a}}\sum_{n\ =\ 1}^{N}{1 \over n^{a}}
+ {1 \over \pars{2N}^{a}}}}
\end{align}
  
  

Then,


*
*When $\ds{a > 1}$ we get
$$
\sum_{n\ =\ 1}^{\infty}{\pars{-1}^{n - 1} \over n^{a}}
=\pars{1 - 2^{1 - a}}\zeta\pars{a}
$$


*However, when $\ds{0\ <\ a\ \leq\ 1}$ 'we'll get a divergence' because
$\ds{\sum_{n\ =\ 1}^{N}{1 \over n^{a}}}$ diverges. In any case, it's true that
$\ds{\lim_{a\ \to\ 0^{+}}\pars{1 - 2^{1 - a}}\zeta\pars{a}=\half}$. In another words, it requires to interpret the sum as proportional to a Riemman Zeta function which involves an analytical continuation.


