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:
$$\eta(s) = \sum_{n=1}^\infty\frac{(-1)^n}{n^s}\;\;,\;\;\;s=\sigma+it\;,\;\;\sigma\,,\,t\in\Bbb R\;,\;\;\color{red}{\sigma > 0}.$$
Now, there seems to be a complete absence 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.
Best Answer
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
$$\frac{t\cos(t\log(x))}{x}$$
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:
$$\left|(2k-1)^{-z} - (2x-1)^{-z}\right| = \left|\int_{2k-1}^{2x-1} \frac{z}{t^{z+1}}\; dt\right| \leq \frac{2|z|}{(2k-1)^{\text{Re}(z) + 1}}.$$
Thus the sum of the integrals is bounded by the finite quantity $2|1+it| \sum_{k=1}^\infty \frac1{(2k-1)^2}$, and we are done.