The integral test applied to $\sum 1/\iota(k+1)^{p}$ for rational $p>0$, cross-checked against the published $p$-series theorem

Let $p \in \mathbb{Q}$ with $p > 0$ (Order on the rationals) and define

$f_p : [0,\infty) \to \mathbb{R}, \qquad f_p(t) \;:=\; (t+1)^{-p} ,$

the rational power of the positive base $t+1 \ge 1$ (Rational powers $a^r$ of a positive base). Then $f_p$ is nonnegative and nonincreasing, so The integral test: for $f \ge 0$ nonincreasing on $[0,\infty)$, $\sum_k f(k)$ converges if and only if the sequence $\bigl(\int_0^N f\bigr)_N$ is bounded, with $\int_0^N f \le \sum_{k<N} f(k) \le f(0) + \int_0^N f$ applies, and its terms are

$f_p(k) \;=\; \frac{1}{\iota(k+1)^{p}} \qquad (k \in \mathbb{N}) .$

The series $\sum_k f_p(k)$ is exactly the $p$-series $\sum_{k \ge 1} 1/\iota(k)^{p}$ in the sense of Series, partial sums, convergence and the sum, divergence, and the tail series, which converges if and only if $p > 1$ (For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$). The integral test therefore delivers, with no primitive computed anywhere:

$\Bigl(\textstyle\int_0^N f_p\Bigr)_{N \in \mathbb{N}} \ \text{ is bounded above} \qquad \Longleftrightarrow \qquad p > 1 .$

The cross-check. At $p = 2$ the integral can also be computed directly: the primitive $G(t) = -(t+1)^{-1}$ gives

$\int_0^N (t+1)^{-2}\,\mathrm{d}t \;=\; 1 - \frac{1}{\iota(N+1)} \;<\; 1 ,$

so the sequence is bounded by $1$, in agreement with the verdict above at $p = 2 > 1$. At $p = 1$ the verdict is that $\bigl(\int_0^N (t+1)^{-1}\bigr)_N$ is unbounded, since the harmonic series diverges. No named logarithmic primitive is available from the current dependency vocabulary, and none is needed for this conclusion.

The exponent must be rational. Real exponents do not exist in this library at this point in the reading order (Why real exponents are deferred on the rational-powers page), so "for $p \in [1,\infty)$" is not a statement that can be made here.

• The test is run backwards here, and that is the point. The usual textbook order computes $\int_1^N t^{-p}\,\mathrm{d}t$ with a primitive and reads off the convergence of the series. That route is unavailable at $p = 1$ in this library, because the primitive is the logarithm and the logarithm is built on a later page. Running the equivalence in the other direction costs nothing: the published For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$ settles the series for every rational $p>0$, and The integral test: for $f \ge 0$ nonincreasing on $[0,\infty)$, $\sum_k f(k)$ converges if and only if the sequence $\bigl(\int_0^N f\bigr)_N$ is bounded, with $\int_0^N f \le \sum_{k<N} f(k) \le f(0) + \int_0^N f$ transfers the verdict to the integrals.

• The index shift is real and is checked in step 2.2. The $p$-series starts at $k = 1$ because $0^{p} = 0$ has no reciprocal, while a sequence in this library is a function on $\mathbb{N}$, which contains $0$; the integrand $(t+1)^{-p}$ is shifted by exactly one for that reason, and $f_p(k) = 1/\iota(k+1)^{p}$. Substituting $t^{-p}$ for $(t+1)^{-p}$ would put an undefined value at $t = 0$ and make $\int_0^N$ improper.

• An independent elementary route to the series verdict. For a nonincreasing nonnegative sequence, $\sum a_k$ converges iff $\sum 2^k a_{2^k}$ converges applies to the nonnegative nonincreasing family $(1/\iota(k)^{p})_{k\ge1}$ and turns it into $\sum_{j\ge0}\iota(2)^{j}/\iota(2^{j})^{p}$, a geometric series; that is the standard elementary route to the same verdict, and it is noted here for orientation only. No claim is made about how For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$ is itself proved, and nothing above depends on this remark.