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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.

user1337
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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 Splaver 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
  • A derivation using the mean value theorem. – Raymond Manzoni Jul 26 '13 at 17:21

2 Answers2

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Hints:

$$\frac1{n^s}-\frac1{(n+1)^s}=s\int\limits_n^{n+1}\frac{dx}{x^{s+1}}$$

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...

DonAntonio
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Here's a direct proof of the statement using Cauchy's criterion given in this file.
Let $K_{\epsilon,M} = \{s\in\Bbb C: Re(s)\geq\epsilon,|s|\leq M\}$ be a compact subset of $Re(s)>0$ for $0<\epsilon\ll 1$ and $1\ll M$. Then, $|n^{-s}|\leq n^{-\epsilon}$ for all $s\in K_{\epsilon,M}$. Let $S_n= \sum_{k=1}^n(-1)^{k+1}$ which is $1$ or $0$. Choose $0<n_1<n_2\in\Bbb N$.

\begin{align*} \left|\sum_{n_1+1}^{n_2}{(-1)^{n+1}\over n^s}\right| & = \left|\sum_{n=n_1+1}^{n_2}{1\over n^s}(S_n-S_{n-1})\right|\\ & = \left|\sum_{n=n_1+1}^{n_2}\left({1\over n^s}-{1\over(n+1)^s}\right)S_n+{S_{n_2}\over(n_2+1)^s}-{S_{n_1}\over(n_1+1)^s}\right|\\ &\leq{1\over(n_1+1)^\epsilon}+{1\over(n_2+1)^\epsilon}+\sum_{n = n_1+1}^{n_2}\left|{1\over n^s}-{1\over(n+1)^s}\right|\\ & \leq {1\over n_1^\epsilon}+{1\over n_2^{\epsilon}}+\sum_{n=n_1+1}^{n_2}\left|\int_n^{n+1}{-s\over x^{s+1}}\ dx\right|\\ & \leq {2\over n_1^{\epsilon}}+\sum_{n=n_1+1}^{n_2}{M\over n^{\epsilon+1}}\\ & \leq {2\over n_1^{\epsilon}}+M\int_{n_1}^{n_2}{1\over x^{1+\epsilon}}\ dx\\ & \leq {2\over n_1^\epsilon}+{M\over\epsilon n_1^{\epsilon}}\to 0,\quad \epsilon\to 0.\\ \end{align*} Hence by Cauchy criterion, the series converges uniformly on a compact subset $K_{\epsilon,M}$ and as $\epsilon,M$ are arbitrary, we conclude the given series defines a holomorphic function on $Re(s)>0$.

one potato two potato
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