A common way of determining that the Riemann zeta function $\sum \frac{1}{n^s}$ converges for $s>1$ is to undertake an integral comparison test.
The following are two examples of this approach:
- Convergence criteria for real Riemann zeta function.
- https://fromprimestoriemann.blogspot.com/2021/02/integral-comparison-tests.html
A key step in the logic is to establish the following inequality, based on the fact that the "area" of the discrete partial sums is less than the area under the related curve:
$$\sum_{n=1}^{k} \frac{1}{n^s} -1 < \int_1^{k}\frac{1}{x^s}\,dx$$
The usual strategy is to find for which $s$ the integral converges as $k \rightarrow \infty$, and then say that the discrete sum also converges for the same condition on $s$, because the sum is always less than the integral.
My question is one about the logic of this strategy:
- the integral converges for $s>1$
- the discrete sum is always less than integral
- so surely the convergence range for the discrete sum can be larger than for the integral?
To expand on this argument - because the sum is always less than the integral, it could be made "larger" by a small amount, and still remain less than the integral. This small enlargement could be done by, $s=0.999" for example.
Where is the flaw in my thinking?
(I am not a university trained mathematician and would welcome replies with minimal technical terminology).