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I would like to know the behavior of the following limit, for $x \in \left(0,\frac{1}{2} \right)$ and $t \in \mathbb{R}$:

$$ L = \lim_{n\to \infty} \dfrac{\sum_\limits{k=1}^n (k + \frac{1}{2})^{x + i\cdot t}}{\sum_\limits{k=1}^n k^{x + i\cdot t}}$$

My observations:

For $t = 0$ one can use Stolz-Cesaro to prove that the limit is $1$. I do not know if there is a limit for $t \neq 0$ Essentially $$ L - 1 = \lim_{n \to \infty} 2^{-x-i\cdot t} \dfrac{\sum_\limits{k=1}^n (-1)^k k^{x+i\cdot t}}{ \sum_\limits{k = 1}^n k ^{x + i\cdot t}}$$ For me would suffice to prove that $|L| \neq 1$ (or $|L| = 1$) or that the limit does not exist

C Marius
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    Asymptotically, for $x\gt0$, $\sum\limits_{k=1}^n\left(k+\frac12\right)^{x+it}\sim\frac1{x+1+it}\left(n+\frac12\right)^{x+1+it}$ and $\sum\limits_{k=1}^nk^{x+it}\sim\frac1{x+1+it}n^{x+1+it}$. – robjohn Oct 10 '20 at 08:31
  • @robjohn that means that $L = 1$ isn't it ? – C Marius Oct 10 '20 at 08:34
  • as long as $\left(1+\frac1{2n}\right)^{x+1+it}\to1$, it would seem so. – robjohn Oct 10 '20 at 08:36
  • @robjohn, in my case for fixed $x \in \left(0, \frac{1}{2}\right)$ and $t > 0$ this is always the case – C Marius Oct 10 '20 at 08:41

1 Answers1

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Using the Euler-Maclaurin Sum Formula, we get $$ \begin{align} \sum_{k=1}^n\left(k+\tfrac12\right)^z &=\frac1{z+1}n^{z+1}+n^z+\zeta(-z)(2^{-z}-1)-2^{-z}+\frac{11z}{24}n^{z-1}+O\!\left(n^{z-2}\right)\tag1\\ \sum_{k=1}^nk^z &=\frac1{z+1}n^{z+1}+\frac12n^z+\zeta(-z)+\frac{z}{12}n^{z-1}+O\!\left(n^{z-3}\right)\tag2 \end{align} $$ The constant in $(1)$ is computed in the same manner as the constant in $(2)$ via analytic continuation. See this answer for details.

Thus, $$ \sum_{k=1}^n\left(k+\tfrac12\right)^z-\sum_{k=1}^nk^z=\frac12n^z+\zeta(-z)(2^{-z}-2)-2^{-z}+\frac{3z}8n^{z-1}+O\!\left(n^{z-2}\right)\tag3 $$ and therefore, for $0\lt\mathrm{Re}(z)\lt1$, $$ \frac{\sum\limits_{k=1}^n\left(k+\tfrac12\right)^z}{\sum\limits_{k=1}^nk^z}=1+\frac{z+1}{2n}+O\!\left(n^{-z-1}\right)\tag4 $$ Finally, we get $$ \lim_{n\to\infty}\frac{\sum\limits_{k=1}^n\left(k+\tfrac12\right)^z}{\sum\limits_{k=1}^nk^z}=1\tag5 $$

robjohn
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