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I want to show that the series $\sum_{n=1}^\infty (-1)^{n+1} n^{-z}$ converges to an analytic function for $\Re z>0$.

For $\Re z>1$ the terms are dominated by $n^{-x}$ so that we have absolute and uniform convergence on compact sets, and by Weierstrass' theorem the sum is analytic there. For $\Re z \leq 1$ however I can't show absolute convergence. I tried splitting into real and imaginary parts: $$\sum_{n=1}^\infty (-1)^{n+1} n^{-z}=\sum_{n=1}^\infty (-1)^{n+1} n^{-x}\cos(-y \ln n)+i\sum_{n=1}^\infty (-1)^{n+1} n^{-x}\sin(-y \ln n),$$ and showing convergence for both using Leibniz's test (or even the more general Dirichlet's test) without success.

I'd love to have any hints about how to do this right.

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$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3‌​^s}+\frac{1}{4^s}..-\frac{2}{2^s}-\frac{2}{4^s}-\frac{2}{6^s}=(1-2^{1-s})\zeta(s)‌​$$ – Ethan Jul 26 '13 at 12:51
@Ethan I believe this representation holds only for $\Re z>1$, where $\zeta$ has its series representation. – user1337 Jul 26 '13 at 12:52
@user1337 The equality stated by Ethan indeed holds for $\mathrm{Re}\,z>0$. I don't think it helps here, though: proving the required properties of $\zeta$ is probably harder than dealing with the alternating sum directly. – 40 votes Jul 26 '13 at 13:17
In the presence of $(-1)^{n+1}$ it is natural to try pairing the terms into $b_k=(2k)^{-s}-(2k+1)^{-s}$. Since the terms of original series go to zero, it converges iff $\sum b_k$ converges. You may be able to get a good estimate for $|b_k|$. – 40 votes Jul 26 '13 at 13:19
See my answer. In this case $a_n=(−1)^n,\, \lambda_n= \log n $ so $A=0$. – user64494 Jul 26 '13 at 16:31
up vote 2 down vote accepted



and now, putting $\,s=\sigma+it\;,\;\;\sigma\,,\,t\in\Bbb R\,$ and taking into account that $\,\sigma>0\,$:

$$\left|\;\int\limits_n^{n+1}\frac{dx}{x^{s+1}}\;\right|\le\int\limits_n^{n+1}\frac{dx}{\left|x^{s+1}\right|}=\int\limits_n^{n+1}\frac{dx}{x^{\sigma+1}}=\left.-\frac1\sigma x^{-\sigma}\right|_n^{n+1}=-\frac1\sigma\left(\frac1{(n+1)^\sigma}-\frac1{n^\sigma}\right)$$

and now observe that

$$-\frac1\sigma\left(\frac1{(n+1)^\sigma}-\frac1{n^\sigma}\right)<\frac1{n^{\sigma+1}}\iff\left[1-\left(\frac n{n+1}\right)^\sigma\right]<1\iff\left(\frac n{n+1}\right)^\sigma>0$$

and now we just apply the series test...

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