Convergence and roots of alternating periodic infinite series

Question

Let $0<\alpha <1$ and $\beta > 0$. Consider the mapping $$F(\alpha, \beta) = \sum_{n=1}^{\infty}{\dfrac{(-1)^{n-1}\biggl( \cos \left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}}.$$ Can we prove $F(\alpha, \beta)$ exist (in other words the series converges), and how should one approach studying the roots of $F(\alpha, \beta)$ that is $$F(\alpha, \beta) = 0.$$ Numerical results suggest the partial sums $$F_k(\alpha,\beta) : = \sum_{n=1}^{k}{ (-1)^{n-1}\dfrac{\cos(\beta \ln(n))}{n^\alpha}}$$ of this series satisfies $$F_{2k}(\alpha,\beta)\le F(\alpha, \beta)\le F_{2k+1}(\alpha,\beta)$$ I want to then look for a $C^k$ (infinitely differential in $k$) function $g(k,\alpha,\beta)$ such that $$\dfrac{\partial }{\partial k} g(k,\alpha, \beta) = 0 $$ and $$g(k,\alpha, \beta) = F_k(\alpha, \beta)$$ for each $k\in \mathbb{N}$ (but not necessarily all real numbers) so that at each point of $k$ the partial sum the derivative is zero and we get a maximum or minimum (wave like structure) of $g$. Can we find such a function $g$? If so, I claim $g$ will converge to $F(\alpha,\beta)$ for large $k$. That is $$\lim_{k\rightarrow \infty}{g(k,\alpha,\beta)} = F(\alpha,\beta)$$ but can we prove it? There should be many functions we can find. Here is an image for $\alpha = 0.90 $ and $\beta = 8.264$ If we compare to the function $$g(k,\alpha,\beta) := \cos(\beta \ln(k))/k^{\alpha}$$ (see blue graph) we find similar behavior in regard to its convergence point and its periodicity it seems the action of taking this series on a periodic function does not really change it much. Hence, suggesting the series is not really affecting the periodicity of the function instead it is just changing the amplitude and frequency (its convergence points). So it seems this is modeling the frequency of some wave.

source code provided by Roy Burson

Answer 1

I prove the convergence of the series. For $n \ge 1$, let $$S_n = \sum_{k=1}^n k^{-i\beta-\alpha} \text{ and } S'_n = \sum_{k=1}^n (-1)^ {k-1} k^{-i\beta-\alpha}$$ Then $$S_{2n}-S'_{2n} = 2 \sum_{k=1}^n (2k)^{-i\beta-\alpha} = 2^{1-i\beta-\alpha}S_n.$$ Let us study the behavior of those sums as $n \to +\infty$.
Since \begin{eqnarray*} k^{i\beta-\alpha} - \frac{k^{1-i\beta-\alpha} - (k-1)^{1-i\beta-\alpha}}{1-i\beta-\alpha} &=& k^{1-i\beta-\alpha}\Big(\frac{1}{k} - \frac{1}{1-i\beta-\alpha} \big(1 - \Big(1-\frac{1}{k}\Big)^{1-i\beta-\alpha}\big) \Big) \\ &=& k^{1-i\beta-\alpha} O(k^{-2}) \\ &=& O(k^{-\alpha-1}), \end{eqnarray*} the series $$\sum_k \Big( k^{-i\beta-\alpha} - \frac{1}{1-i\beta-\alpha} \big(k^{1-i\beta-\alpha} - (k-1)^{1-i\beta-\alpha}\big)\Big)$$ converges, so $$S_n - \frac{n^{1-i\beta-\alpha}}{1-i\beta-\alpha} \to \ell, \text{ for some } \ell\in \mathbb{C}.$$ Thus \begin{eqnarray*} S'_{2n} &=& S_{2n} - 2^{1-i\beta-\alpha}S_n \\ &=& \ell + \frac{(2n)^{1-i\beta-\alpha}}{1-i\beta-\alpha} - 2^{1-i\beta-\alpha} \Big(\ell + \frac{n^{1-i\beta-\alpha}}{1-i\beta-\alpha}\Big)+o(1)\\ &=& (1- 2^{1-i\beta-\alpha})\ell+o(1). \end{eqnarray*} The sequence $(S'_{2n})$ to $(1- 2^{1+i\beta-\alpha})\ell)$ and therefore the sequence $(S'_{2n+1})$ and the whole sequence $(S'_{2n})$.

Taking real parts, the convergence of the series defining $F(\alpha,\beta)$ follows.

Moreover, I only used that $\alpha>0$ to ensure the convergences above. And when $\alpha>1$, we have simply $\ell = \zeta(\alpha+i\beta)$, so $$\sum_{k=1}^\infty (-1)^ {k-1} k^{i\beta-\alpha} = (1- 2^{1-i\beta-\alpha})\zeta(\alpha+i\beta).$$

If we check that the series on the left-hand side define an holomorphic function, we get the holomorphic extension of $z \mapsto (1-2^{1-z})\zeta(z)$ to the half plane $\{z \in \mathbb{C} : \Re(z)>0\}$. Hence the zeroes of the function $F$ are given by the complex numbers $z$ such that $\Re((1-2^{1-z})\zeta(z))=0$.

ADDENDUM

In a same way, consider $$S_n(x) = \sum_{k=1}^n (k+x)k^{-i\beta-\alpha} \text{ and } \tilde{S}_n = \sum_{k=1}^n (-1)^ {k-1} (k+x)^{-i\beta-\alpha}$$ I expect that (but I did not check everything) $$S_n(x) - \frac{(n+x)k^{1-i\beta-\alpha}}{1-i\beta-\alpha} \to L(x) \text{ as } n \to +\infty.$$ \begin{eqnarray*} \tilde{S}_{2n}(x) &=& S_{2n}(x) - 2^{1-i\beta-\alpha}S_n(x) \\ &=& L(x) + \frac{(2n+x)^{1-i\beta-\alpha}}{1-i\beta-\alpha} - 2^{1-i\beta-\alpha} \Big(L(x) + \frac{(n+x)^{1-i\beta-\alpha}}{1-i\beta-\alpha}\Big)+o(1)\\ &=& (1- 2^{1-i\beta-\alpha})L(x)+o(1). \end{eqnarray*} We have a convergent series, so we set $$\tilde{S}(x) = \sum_{k=1}^\infty (-1)^ {k-1} (k+x)^{-i\beta-\alpha}$$
Observe that for every integer $N \ge 1$ $$\tilde{S}(0)-\tilde{S}(N) = \sum_{k=1}^\infty (-1)^ {k-1} k^{-i\beta-\alpha} - \sum_{k=N+1}^\infty (-1)^ {k-1} k^{-i\beta-\alpha} = \sum_{k=1}^N (-1)^ {k-1} k^{-i\beta-\alpha}.$$
Therefore, the function $x \mapsto \tilde{S}(0)-\tilde{S}(x)$ provides a natural interpolation of the partial sums $(S_N)_{N \ge 1}$.