I'd like to check directly the convergence of Dirichlet's Eta Function, also known as the Alternating Zeta Function or even Alternating Euler's Zeta Function:

$$\sum_{n=1}^\infty\frac1{n^s}\;\;,\;\;\;s=\sigma+it\;,\;\;\sigma\,,\,t\in\Bbb R\;,\;\;\color{red}{\sigma > 0}.$$

Now, there seems to be a complete ausence of any direct proof of this in the web (at least I didn't find it) that doesn't use the theory of general Dirichlet Series and things like that.

I was thinking of the following direct, more elementary approach:

$$n^{it}=e^{it\log n}:=\cos (t\log n)+i\sin(t\log n)$$

and then we can write

$$\frac1{n^s}=\frac1{n^\sigma n^{it}}=\frac{\cos(t\log n)-i\sin(t\log n)}{n^\sigma}$$

and since a complex sequence converges iff its real and imaginary parts converge, we're left with the real series

$$\sum_{n=1}^\infty\frac{\cos(t\log n)}{n^\sigma}\;\;,\;\;\;\;\sum_{n=1}^\infty\frac{\sin(t\log n)}{n^\sigma}$$

Now, I think it is enough to prove only one of the above two series' convergence, since for example $\;\sin(t\log n)=\cos\left(\frac\pi2-t\log n\right)\;$

...and here I am stuck. It seems obvious both series are alternating but not necessarily elementwise.

For example, if $\;t=1\;$ , then $\;\cos\log n>0\;,\;for\;\;n=1,2,3,4\;$ , and then $\;\cos\log n<0\;,\;\;for\;\;\;n=5,6,\ldots,23\;$ . This behaviour confuses me, and any help will be much appreciated.

  • $\begingroup$ Apostol's "Introduction to Analytic Number Theory" covers this. $\endgroup$ – Pedro Tamaroff Dec 8 '14 at 5:40
  • $\begingroup$ I would split the sums into the regions when $\cos(t \log n)$ (or $\sin(t \log n)$) are positive and negative and estimate the sums in each region. You have started to do this. $\endgroup$ – marty cohen Dec 8 '14 at 6:16
  • $\begingroup$ @PedroTamaroff Thank you very much, yet I cannot find it in the book. There is a two lines reference to this function in 11.6 (page 234, Ed. Springer 1976), yet it only says the function converges for Re$(s)>0\;$ , without any proof, at least there. Perhaps it is mentioned and proved somewhere before this page? $\endgroup$ – user177692 Dec 9 '14 at 3:41
  • $\begingroup$ Apostol might have references at the end of the book, too. $\endgroup$ – Pedro Tamaroff Dec 9 '14 at 3:46
  • $\begingroup$ Your definition of $\eta$ seems to be missing a $(-1)^n$ ... $\endgroup$ – Pixel Jul 12 '17 at 21:23

I think this succumbs to applications of Dirichlet's test and some estimates based on the mean value theorem.

Dirichlet's test gives conditional convergence of a sum $\sum_{n \geq 1} a_n b_n$ provided that $a_n$ is a sequence of positive numbers that decrease to $0$ and the sequence of partial sums $B_n = \sum_{k=1}^n b_k$ is uniformly bounded in absolute value. In this case, take $a_n = n^{-\sigma}$ and $b_n = (-1)^{n+1} n^{-it}$. Clearly the $a_n$ decrease to $0$ if $\sigma > 0$, so we just need to get the uniform boundedness of the $B_n$.

It suffices to bound $\sum_{k=1}^n (2k-1)^{-it} - (2k)^{-it}$. Clearly this is bounded for $t=0$, so assume $t\neq 0$. The real and imaginary parts are

$$C_n = \sum_{k=1}^n \cos(t\log(2k-1)) - \cos(t\log(2k)),$$

$$S_n = \sum_{k=1}^n \sin(t\log(2k)) - \sin(t\log(2k-1)).$$

Let's show how to bound the imaginary part $S_n$; the real part is bounded by a wholly analogous technique.

By the mean value theorem, the $k$-th term of $S_n$ is


for some $x \in [2k-1, 2k]$. This is not much different from $\frac{t\cos(t\log(2k-1))}{2k-1}$; in fact, the difference between them is, again by the mean value theorem,

$$\frac{-t^2 \sin(t\log(y)) +t\cos(t\log(y))}{y^2}(x - (2k-1))$$

for some $y \in [2k-1, x]$, whose absolute value is bounded above by $\frac{t^2+t}{(2k-1)^2}$. Since $\sum_{k\geq 1} \frac1{(2k-1)^2}$ converges, this reduces us to putting a uniform bound on the absolute value of

$$\sum_{k=1}^n \frac{\cos(t\log(2k-1))}{2k-1} = \text{Re}\sum_{k=1}^n \frac1{(2k-1)^{1 + it}}.$$

Put $z = 1 + it$; we may as well bound the absolute value of

$$\sum_{k=1}^n \frac1{(2k-1)^z} = \sum_{k=1}^n \left[\int_k^{k+1} \frac1{(2k-1)^z} - \frac1{(2x-1)^z} dx\right] + \int_1^{n+1} \frac1{(2x-1)^z} dx.$$

The last integral is easily evaluated as $\left[\frac{(2x-1)^{-it}}{-2it}\right]_1^{n+1}$ which in absolute value is bounded above by $\frac1{t}$. We turn now to the preceding sum of integrals. The integrands are bounded thus:

$$\left|(2k-1)^{-z} - (2x-1)^{-z}\right| \leq \max_{k \leq y \leq k+1} \left|(-2z)(2y-1)^{-2 - it}\right| \leq (2+2t)(2k-1)^{-2}$$

where we again used the mean value theorem in the first inequality. Thus the sum of the integrals is bounded by the finite quantity $(2 + 2t) \sum_{k=1}^\infty \frac1{(2k-1)^2}$, and we are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy