Question 1:
I have recently tried out answering a couple of questions with the method of using the same variable in the argument of a function and a sum for example:
$$\int_0^N (\sqrt{x+1}-\sqrt x)^n dx= \lim_{N\to ∞}\sum_{k=0}^N f\biggl(a+k\frac{b-a}{N}\biggl )\frac{b-a}{N}= \lim_{N\to ∞}\sum_{k=0}^N f\biggl(0+k\frac{N-0}{N}\biggl )\frac{N-0}{N}= \lim_{N\to ∞}\sum_{k=0}^N f(k)*1= \lim_{N\to ∞}\sum_{k=0}^N (\sqrt{k+1}-\sqrt k)^n= \sum_{k=0}^ ∞(\sqrt{k+1}-\sqrt k)^n$$
Therefore, with a restriction: $$\lim_{N\to\infty}\bigg(\int_0^N (\sqrt{x+1}-\sqrt x)^n dx= \sum_{k=0}^ N(\sqrt{k+1}-\sqrt k)^n\bigg),1\ge n\in \Bbb R$$ Here is proof of the answer.
This works and even shows that the integral or sum operator does not matter up to a point. Why does this work with the same operators and sum index bound in the integral bound?
Question 2:
This is the main reason I made this question. Imagine with an easier example that: $$S=\sum_{n\ge 1} 2^{-n} $$ This simple geometric series looks like the reimann zeta function, but with the wrong order of exponentiation where x is a function to be determined: $$\sum_{n\ge 1} 2^{-n}=ζ(x)=\sum_{n\ge 1}n^{-x} $$
To solve for x, equate the like expressions: $$2^{-n}= n^{-x}⇒x=-\log_n(2^{-n})=n\log_n(2)$$ Therefore n is in the argument of the Riemann Zeta function as $x(n)$: $$S=ζ(n\log_n(2))= \sum_{n\ge 1}2^{-n}, \mathrm {restrictions}$$
However, this obviously does not work unless x(n) is plugged into the Riemann Zeta definition. What would this mean, if anything, to have an argument as an already defined index of a function and why does this fail? Please give me feedback and correct me!
Sorry if this question is trivial. I am genuinely curious about this peculiar property and you may be as well.