How the nonnegative tests are ordered by strength, and which of them this page cannot state without the logarithm

The tests on this page are not independent criteria of comparable status. Some of them are strictly stronger than others, in the precise sense that whenever the weaker one decides a series, the stronger one decides it the same way, and there are series the stronger one decides and the weaker one does not. This remark records exactly which comparisons are proved here, and, equally importantly, which are not.

Everything on this page is a comparison in disguise. If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$, and divergence of $\sum a_k$ gives divergence of $\sum b_k$ compares against an arbitrary series; the strength of every later test is the strength of the particular series it compares against. The root and ratio tests compare against a geometric series (For $|r| < 1$, $\sum_{k \ge 0} r^k = 1/(1-r)$, and for $|r| \ge 1$ the series diverges); Raabe's test compares against the harmonic series, through the weights $\zeta_k = k+1$ in Kummer: for positive terms $a_k$ and weights $\zeta_k > 0$, $\liminf(\zeta_k a_k/a_{k+1} - \zeta_{k+1}) > 0$ gives convergence, and if $\sum 1/\zeta_k$ diverges while that expression is eventually $\le 0$ the series diverges; and the borderline branch of Gauss: for positive terms, if $a_k/a_{k+1} = 1 + h/k + r_k$ with $|r_k| \le C\,k^{-1-\varepsilon}$ for $k \ge 1$, some constant $C$ and some rational $\varepsilon > 0$, the series converges iff $h > 1$ compares against the harmonic series again. For a nonincreasing nonnegative sequence, $\sum a_k$ converges iff $\sum 2^k a_{2^k}$ converges is of a different kind: it does not compare, it reindexes, and that is why it settles the whole $p$-series family (For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$) in one step.

The comparisons proved on this page.

• Root over ratio. Whenever the ratio test decides, the root test decides the same way, and the converse fails: whenever the ratio test (Ratio test: $\limsup |a_{k+1}/a_k| < 1$ gives absolute convergence and hence convergence, and $\liminf |a_{k+1}/a_k| > 1$ gives divergence) decides, the root test (Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing) decides the same way, and there is a series the root test decides and the ratio test does not. This is a consequence of the inequality chain $\liminf q_k \le \liminf \rho_k \le \limsup \rho_k \le \limsup q_k$, proved on the previous page, and of nothing else.
• Ratio as a case of Kummer. Kummer with $\zeta_k = 1$ recovers the ratio test: for positive terms, the ratio test is Kummer's test with the constant weights $\zeta_k = 1$.
• Raabe as a case of Kummer. Raabe is Kummer with $\zeta_k = k+1$: for positive terms, $\liminf\, (k+1)(a_k/a_{k+1} - 1) > 1$ gives convergence and $\limsup < 1$ gives divergence: Raabe's test is Kummer's test with the weights $\zeta_k = k+1$.
• Gauss over Raabe, inside Gauss's hypothesis. Gauss: for positive terms, if $a_k/a_{k+1} = 1 + h/k + r_k$ with $|r_k| \le C\,k^{-1-\varepsilon}$ for $k \ge 1$, some constant $C$ and some rational $\varepsilon > 0$, the series converges iff $h > 1$ assumes an expansion $a_k/a_{k+1} = 1 + h/k + r_k$ with a summable error. Under that hypothesis $R_k = (k+1)(a_k/a_{k+1}-1)$ converges to $h$. When $h \ne 1$, Raabe's test already decides, and Gauss's proof says so by invoking it. When $h = 1$ Raabe's test cannot decide: $\liminf_k R_k > 1$ would force $R_k$ to stay above a fixed number greater than $1$ from some index on, and $\limsup_k R_k < 1$ would force it to stay below a fixed number less than $1$, and $R_k \to 1$ rules out both. Gauss decides that case, and it is the reason the theorem exists.

Two comparisons that are not claimed here. Raabe's test is not compared with the root test on this page, in either direction, and nothing above should be read as ordering them. Nor is Kummer's test claimed to be universal: the choice of weights is free, and the question of which series some choice of weights decides is not addressed.

And one that is refuted. No comparison test can be final. For a divergent series of positive terms with partial sums $s_k$, the series $\sum a_k/s_k$ diverges and $\sum a_k/s_k^2$ converges turns any divergent series of positive terms into a divergent series of positive terms with eventually smaller terms, and FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test draws the conclusion: there is no slowest divergent series of positive terms, hence no universal comparison test. The hierarchy above is therefore an initial segment of something with no last term, not an approach to a best test.

What this page cannot state, and why. Every gap below is a missing definition, not a missing proof.

• The $p$-series at irrational exponents. Rational powers $a^r$ of a positive base defines $a^{r}$ for rational $r$ and positive $a$. So $\sum 1/k^{p}$ is a well-formed expression here only for rational $p$, and For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$ is the full theorem for every exponent this page can name. Real exponents wait for the exponential and the logarithm.
• Bertrand's test. Its criterion is a condition on $\log k \,\big(k(a_k/a_{k+1} - 1) - 1\big)$, and it is the natural next member of the Kummer family, with weights $\zeta_k = k \log k$. Both the weights and the criterion mention the logarithm, so neither can be written down here.
• The integral test. It compares $\sum f(k)$ with $\int f$, and the Riemann integral is developed much later in this library. Condensation is the substitute used on this page, and for the $p$-series it does the same work.
• The general form of Gauss's test. The classical statement assumes $r_k=O(k^{-\beta})$ for some real $\beta>1$. The version proved here writes $\beta=1+\varepsilon$ with $\varepsilon$ a positive rational. This loses no case covered by the classical hypothesis: given $\beta>1$, choose a rational $0<\varepsilon<\beta-1$ and weaken the eventual bound. An error of order $1/(k\log k)$ is not a Gauss remainder at $h=1$; it is the next Bertrand borderline.

A limitation that has been removed, and one that has not. The comparison with a geometric series inside Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing and Ratio test: $\limsup |a_{k+1}/a_k| < 1$ gives absolute convergence and hence convergence, and $\liminf |a_{k+1}/a_k| > 1$ gives divergence delivers convergence of $\sum |a_k|$ and not, on its own, of $\sum a_k$. That the second follows from the first is If $\sum |a_k|$ converges then $\sum a_k$ converges, proved on this page from A series converges iff for every $\varepsilon > 0$ there is $N$ with $|a_{m+1} + \dots + a_n| < \varepsilon$ for all $n > m \ge N$ and the triangle inequality for finite sums, so both tests do reach their standard conclusion here. What is not on this page is the rest of that theory: the converse fails, and the alternating harmonic series that witnesses the failure needs the alternating series test, which is not proved here; rearrangement, the Riemann series theorem and products of series belong with it on the page that follows. Nothing above asserts a converse or identifies any sum.