Compute this complex limit I am having difficulties in proving that the below limit exists:
$$\LARGE{ \lim_{n \to \infty} \frac{\sum_{k=1}^n \frac{1}{(k+x)^{x + i\cdot t}}}{\sum_{k=1}^n \frac{1}{k^{x + i\cdot t}}}}$$ where $x \in (0,1)$ and $t \in \mathbb{R}_+$
I would like to have an upper bound on the modulus of this thing, for instance as a function of $t,x$.
My observations:
for $x = 0$ the above is $1$, for $t = 0$ one can use Stolz-Cesaro and prove that the limit is $ \lim_{n \to \infty} \left( \frac{n}{n+x} \right)^{x} = 1$. For $x = 1$ one obtains something like $\lim_{n \to \infty} \frac{S_n - 1}{S_n}$ where $S_n$ is the denominator of the above limit. Again, if $|S_n| \to \infty$ then the limit apporoaches $1$ in modulus. How can I prove this for the general case? If the function is holomorphic, then can we use the identity theorem for holomorphic functions to say that the limit is $1$ for all $t$ since it is $1$ for all $\{x + i \cdot t | t = 0, x \in (0,1) \}$ (which has an accumulation point)
For $x = 0$: $\lim _{n\rightarrow \infty }\frac{\sum\limits ^{n}_{k=1}\frac{1}{k^{it}}}{\sum\limits ^{n}_{k=1}\frac{1}{k^{it}}} =1$.
For $x = 1$: $\lim _{n\rightarrow \infty }\frac{\sum\limits ^{n}_{k=1}\frac{1}{( k+1)^{1+it}}}{\sum\limits ^{n}_{k=1}\frac{1}{k^{1+it}}} =\lim _{n\rightarrow \infty }\frac{\frac{1}{2^{1+it}} +\frac{1}{3^{1+it}} +\frac{1}{4^{1+it}} +\cdots }{\frac{1}{1^{1+it}} +\frac{1}{2^{1+it}} +\frac{1}{3^{1+it}} +\cdots } = 1$, because of $\lim _{n \to \infty} \frac{1^t + 2^t + 3^t + \cdots}{2^t + 3^t + \cdots} = 1$.
Then $\lim _{n\rightarrow \infty }\frac{\sum\limits ^{n}_{k=1}\frac{1}{( k+x)^{x+it}}}{\sum\limits ^{n}_{k=1}\frac{1}{k^{x+it}}} = 1$.
For $x = 0$: $\lim _{n\rightarrow \infty }\frac{\sum\limits ^{n}_{k=1}\frac{1}{k^{it}}}{\sum\limits ^{n}_{k=1}\frac{1}{k^{it}}} =1$.
For $x = 1$: $\lim _{n\rightarrow \infty }\frac{\sum\limits ^{n}_{k=1}\frac{1}{( k+1)^{1+it}}}{\sum\limits ^{n}_{k=1}\frac{1}{k^{1+it}}} =\lim _{n\rightarrow \infty }\frac{\frac{1}{2^{1+it}} +\frac{1}{3^{1+it}} +\frac{1}{4^{1+it}} +\cdots }{\frac{1}{1^{1+it}} +\frac{1}{2^{1+it}} +\frac{1}{3^{1+it}} +\cdots } = 1$, because of $\lim _{n \to \infty} \frac{1^t + 2^t + 3^t + \cdots}{2^t + 3^t + \cdots} = 1$.
Then $\lim _{n\rightarrow \infty }\frac{\sum\limits ^{n}_{k=1}\frac{1}{( k+x)^{x+it}}}{\sum\limits ^{n}_{k=1}\frac{1}{k^{x+it}}} = 1$.