# Showing a series based on the Riemann zeta function converges (uniformly)

This question comes from Lars Ahlfors' complex analysis (page 178).

We define $\zeta(z) = \sum_{n=1}^{\infty} n^{-z}$. This is just the Riemann zeta function. I am struggling however to prove that the series

$$(1-2^{1-z})\zeta(z) = \sum_{n=1}^{\infty} (-1)^{n+1}n^{-z}$$ will converge whenever $\Re (z) >0.$

As far as I can tell, the alternating series test does not apply. I have attempted to use Dirichlet's Test, but to no avail.

I also have tried looking at the real and imaginary parts of the series.

What are some techniques I could use to show this series converges?

Sketch: Let $S_n =\sum_{m=1}^{n}(-1)^m.$ Then, summing by parts, we get

$$\sum_{n=1}^{N}(-1)^nn^{-z} = S_N N^{-z} + \sum_{n=1}^{N-1}S_n[n^{-z}-(n+1)^{-z}].$$

The first term on the right $\to 0.$ So we'll be done if we show

$$\sum_{n=1}^{\infty}|n^{-z}-(n+1)^{-z}| < \infty.$$

I'm seeing the $n$th term of the last series as on the order of $1/n^{1+\text {Re} z}.$

We have $$\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{n^{z}}=\sum_{n\geq1}\frac{1}{\left(2n-1\right)^{z}}-\frac{1}{\left(2n\right)^{z}}=\sum_{n\geq1}\int_{2n-1}^{2n}\frac{z}{x^{z+1}}dx$$ hence $$\left|\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{n^{z}}\right|\leq\left|z\right|\int_{1}^{\infty}\frac{1}{x^{\textrm{Re}\left(z\right)+1}}dx$$ which converges if $\textrm{Re}\left(z\right)>0$.

• Is the first equality valid without absolute convergence? Nov 9 '19 at 17:09

$\displaystyle\frac{1}{n^s} - \frac{1}{(n+1)^s} = \frac{1}{n^s} \left(1 - \left(1 - \frac{1}{n+1}\right)^s \right)$. look now at the function $\displaystyle f(x) = \left(1 + x\right)^s$. it's $C^\infty$ at $x \approx 0$ and $f'(x) = s \left(1 + x\right)^{s-1}$ so that $\displaystyle f\left(-\frac{1}{n+1}\right) = 1 - \frac{s}{n+1} + \mathcal{O}\left(\frac{1}{(n+1)^2}\right)$ and $\displaystyle\frac{1}{n^s} - \frac{1}{(n+1)^s} = \frac{s}{n^s(n+1)} + \mathcal{O}(n^{-s-2})$ and finally :

$$\sum_{n=1}^\infty \frac{1}{(2n-1)^{s}} - \frac{1}{(2n)^s} = \sum_{n=1}^\infty \frac{s}{2n(2n-1)^s} + \mathcal{O}(n^{-s-2})$$

but the most important thing is that : given a Dirichlet series, if we know $\sum a_n n^{-s}$ converges, then for any $z$ of the right half plane $\Re(z) > \Re(s)$ the series $\sum a_n n^{-z}$ converges. this means you have to find only one value for $s$ where the series converges, then it will converge in all the half plane on the right of it.

you can prove it with summation by parts and that $\displaystyle\frac{1}{n^s} - \frac{1}{(n+1)^s} = \mathcal{O}(n^{-s-1})$.

personnally I learnt it in a book named "General theory of Dirichlet series" where the formula for the abscissa of convergence of a Dirichlet series is explained, the equivalent of radius of convergence for a Taylor series.