Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3‌​^s}+\frac{1}{4^s}..-\frac{2}{2^s}-\frac{2}{4^s}-\frac{2}{6^s}=(1-2^{1-s})\zeta(s)‌​$$ –  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
show 1 more comment

1 Answer 1

up vote 2 down vote accepted

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...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.