zeros of Dirichlet eta function

As stated in the parent entry (http://planetmath.org/AnalyticContinuationOfRiemannZetaToCriticalStrip), the definition of the Riemann zeta function may be analytically continued (http://planetmath.org/AnalyticContinuation) from the half-plane  $\mathrm{\Re }s>1$  to the half-plane  $\mathrm{\Re }s>0$  by using the Dirichlet eta function $\eta (s)$ via the equation

$\zeta (s)={\displaystyle \frac{\eta (s)}{1-\frac{2}{2^s}}}.$(1)

Then only the status of the points

$s_n:=\mathrm{\hspace{0.33em}1}+n\cdot {\displaystyle \frac{2\pi i}{\ln 2}}\mathit{\hspace{1em}\hspace{1em}}(n\in \mathbb{Z})$(2)

which are the zeros of $1-\frac{2}{2^s}$, remains :  are they poles of $\zeta (s)$ or not?  E. Landau has 1909 signaled this problem, which has been elementarily solved not earlier than after 40 years, by D. V. Widder.  He proved that those numbers, except  $s=1$,  are also zeros of $\eta (s)$.  This means that they only are removable singularities of $\zeta (s)$ and that (1) in fact extends $\zeta (s)$ to every points of the half-plane $\mathrm{\Re }s>0$  except  $s=1$.

A new direct proof by J. Sondow of the vanishing of the Dirichlet eta function at the points  $s_n\ne 1$  was published in 2003.  It is based on a relation between the partial sums ${\eta }_n(s)$ and${\zeta }_n(s)$ of the series defining respectively the functions $\eta (s)$ and $\zeta (s)$ for $\mathrm{\Re }s>1$, which involves the approximation of an integral by a Riemann sum.

With some clever but not so complicated performed on finite sums, Sondow writes for any $s$ the following:

${\eta }_{2n}(s)$$\mathrm{\hspace{0.33em}}=\mathrm{\hspace{0.33em}1}-{\displaystyle \frac{1}{2^s}}+{\displaystyle \frac{1}{3^s}}-{\displaystyle \frac{1}{4^s}}+-\mathrm{\dots }+{\displaystyle \frac{{(-1)}^{2n-1}}{{(2n)}^s}}$

$\mathrm{\hspace{0.33em}}=\mathrm{\hspace{0.33em}1}+{\displaystyle \frac{1}{2^s}}+{\displaystyle \frac{1}{3^s}}+{\displaystyle \frac{1}{4^s}}+\mathrm{\dots }+{\displaystyle \frac{{(-1)}^{2n-1}}{{(2n)}^s}}-2\left({\displaystyle \frac{1}{2^s}}+{\displaystyle \frac{1}{4^s}}+\mathrm{\dots }+{\displaystyle \frac{1}{{(2n)}^s}}\right)$

$\mathrm{\hspace{0.33em}}=\left(1-{\displaystyle \frac{2}{2^s}}\right){\zeta }_{2n}(s)+{\displaystyle \frac{2}{2^s}}\left({\displaystyle \frac{1}{{(n+1)}^s}}+\mathrm{\dots }+{\displaystyle \frac{1}{{(2n)}^s}}\right)$

$\mathrm{\hspace{0.33em}}=\left(1-{\displaystyle \frac{2}{2^s}}\right){\zeta }_{2n}(s)+{\displaystyle \frac{2n}{{(2n)}^s}}{\displaystyle \frac{1}{n}}\left({\displaystyle \frac{1}{{(1+1/n)}^s}}+\mathrm{\dots }+{\displaystyle \frac{1}{{(1+n/n)}^s}}\right)$

Now if $t$ is real,  $s=1+it$,  and  $2^{1-s}=2^{-it}=1$,  then the factor multiplying ${\zeta }_{2n}(s)$ is zero and consequently

$${\eta }_{2n}(s)=\frac{1}{n^{it}}{R}_n(1/{(1+x)}^s,0,1)$$

where  $R_n(f(x),a,b)$  denotes a special Riemann sum approximating the integral of $f(x)$ over  $[a,b]$.  For  $s=1$,  i.e.  $t=0$, one gets

$$\eta (1)=\underset{n\to \mathrm{\infty }}{lim}{\eta }_{2n}(1)=\underset{n\to \mathrm{\infty }}{lim}{R}_n(1/(1+x),0,1)={\int }_0^1\frac{dx}{1+x}=\ln 2$$

and otherwise, when  $t\ne 0$,  one has  $|n^{1-s}|=|n^{-it}|=1$,  giving

$|\eta (s)|$$\mathrm{\hspace{0.33em}}=\underset{n\to \mathrm{\infty }}{lim}|{\eta }_{2n}(s)|=\underset{n\to \mathrm{\infty }}{lim}|R_n(1/{(1+x)}^s,0,1)|$

$\mathrm{\hspace{0.33em}}=\left|{\displaystyle {\int }_0^1}{\displaystyle \frac{dx}{{(1+x)}^s}}\right|=\left|{\displaystyle \frac{2^{1-s}-1}{1-s}}\right|=\left|{\displaystyle \frac{1-1}{-it}}\right|=\mathrm{\hspace{0.33em}0}.$

Note.  By (1) the Dirichlet eta function has as zerosalso the zeros of the Riemann zeta function (seeRiemann hypothesis (http://planetmath.org/RiemannZetaFunction)).