How to compute $\lim\limits_{s\rightarrow0^{+}}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{s}}$

Question

Recently I have been discussing the question with my classmates. Since the partial sum of $(-1)^{n}$ is uniformly bounded and $\frac{(-1)^{n}}{n^{s}}$ decreases to $0$ uniformly on any closed interval in $(0,+\infty)$, we know $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{s}}$ is continuous on $(0,+\infty)$. However, we found it diffucult to compute $\lim\limits_{s\rightarrow0^{+}}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{s}}$. I am not sure whether we can use the Riemann-zeta function $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}$ since the definition of $\zeta(s)$ around $0$ depends on the analytic extension instead of the sum itself. Any advice will be helpful.

Answer 1

Notice that for any $s>0$ $$-\eta(s)=\sum_{n=1}^\infty\frac{(-1)^n}{n^s}=\sum_{n=1}^\infty\frac{(-1)^n}{\Gamma(s)}\int_0^\infty t^{s-1}e^{-nt}dt=\frac{1}{\Gamma(s)}\lim_{N\to\infty}\sum_{n=1}^N\int_0^\infty (-1)^nt^{s-1}e^{-nt}dt$$ Next, consider for any fixed $N$ the set of functions $f_N(t)=\sum_{n=1}^N(-1)^ne^{-nt}t^{s-1}=\frac{-1+(-1)^{N+1}e^{-Nt}}{e^t+1}t^{s-1}$

$|f_N(t)|$ is dominated by the integrable function $f(t)=\frac{2t^{s-1}}{e^t+1}$ on $[0;\infty)$. Therefore, by the dominated convergence theorem (DCT), $$\lim_{N\to\infty}\int_0^\infty f_N(t)dt=\int_0^\infty\Big(\lim_{N\to\infty}f_N(t)\Big)dt$$ and $$-\eta(s)=\frac{1}{\Gamma(s)}\lim_{N\to\infty}\sum_{n=1}^N(-1)^n\int_0^\infty t^{s-1}e^{-nt}dt=\frac{1}{\Gamma(s)}\int_0^\infty \Big(\lim_{N\to\infty}\sum_{n=1}^N(-1)^nt^{s-1}e^{-nt}\Big)dt$$ $$=-\frac{1}{\Gamma(s)}\int_0^\infty\frac{t^{s-1}}{e^t+1}dt\overset{IBP}{=}-\,\frac{1}{s\Gamma(s)}\frac{t^s}{e^t+1}\,\bigg|_0^\infty-\,\frac{1}{s\Gamma(s)}\int_0^\infty\frac{t^se^t}{(e^t+1)^2}dt$$ $$-\eta(s)=-\frac{1}{\Gamma(1+s)}\int_0^\infty\frac{t^se^t}{(e^t+1)^2}dt$$ Leading $s\to 0$, we can use again the DCT, and take the limit under the integral sign. Given that $\lim_{s\to 0}\Gamma(1+s)=1$, $$\lim_{s\to0^+}\big(-\eta(s)\big)=-\,\int_0^\infty\frac{e^t}{(e^t+1)^2}dt=-\,\frac{1}{2}$$

Answer 2

$$f(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{s}}=\left(2^{1-s}-1\right) \zeta (s)=-\eta(s)$$ Expanded as a series around $s=0^+$ (have look here) $$f(s)=-\frac{1}{2}+\frac{1}{2} \log \left(\frac{2}{\pi }\right)s+O\left(s^2\right)$$ which is quite good up to $s\sim \frac 14$

$$\int_0^{\frac 14}\Bigg[\left(2^{1-s}-1\right) \zeta (s)-\left(-\frac{1}{2}+\frac{1}{2} \log \left(\frac{2}{\pi }\right)s \right)\Bigg]^2\,ds=1.906\times 10^{-7}$$