Proof of Raabe's test

Defining ${\displaystyle \rho _{n}\equiv n\left({\frac {a_{n}}{a_{n+1}}}-1\right)}$, we need not assume the limit exists; if ${\displaystyle \limsup \rho _{n}<1}$, then ${\displaystyle \sum a_{n}}$ diverges, while if ${\displaystyle \liminf \rho _{n}>1}$ the sum converges.

The proof proceeds essentially by comparison with ${\displaystyle \sum 1/n^{R}}$. Suppose first that ${\displaystyle \limsup \rho _{n}<1}$. Of course if ${\displaystyle \limsup \rho _{n}<0}$ then ${\displaystyle a_{n+1}\geq a_{n}}$ for large ${\displaystyle n}$, so the sum diverges; assume then that ${\displaystyle 0\leq \limsup \rho _{n}<1}$. There exists ${\displaystyle R<1}$ such that ${\displaystyle \rho _{n}\leq R}$ for all ${\displaystyle n\geq N}$, which is to say that ${\displaystyle a_{n}/a_{n+1}\leq \left(1+{\frac {R}{n}}\right)\leq e^{R/n}}$. Thus ${\displaystyle a_{n+1}\geq a_{n}e^{-R/n}}$, which implies that ${\displaystyle a_{n+1}\geq a_{N}e^{-R(1/N+\dots +1/n)}\geq ca_{N}e^{-R\log(n)}=ca_{N}/n^{R}}$ for ${\displaystyle n\geq N}$; since ${\displaystyle R<1}$ this shows that ${\displaystyle \sum a_{n}}$ diverges.

The proof of the other half is entirely analogous, with most of the inequalities simply reversed. We need a preliminary inequality to use in place of the simple ${\displaystyle 1+t<e^{t}}$ that was used above: Fix ${\displaystyle R}$ and ${\displaystyle N}$. Note that ${\displaystyle \log \left(1+{\frac {R}{n}}\right)={\frac {R}{n}}+O\left({\frac {1}{n^{2}}}\right)}$. So ${\displaystyle \log \left(\left(1+{\frac {R}{N}}\right)\dots \left(1+{\frac {R}{n}}\right)\right)=R\left({\frac {1}{N}}+\dots +{\frac {1}{n}}\right)+O(1)=R\log(n)+O(1)}$; hence ${\displaystyle \left(1+{\frac {R}{N}}\right)\dots \left(1+{\frac {R}{n}}\right)\geq cn^{R}}$.

Suppose now that ${\displaystyle \liminf \rho _{n}>1}$. Arguing as in the first paragraph, using the inequality established in the previous paragraph, we see that there exists ${\displaystyle R>1}$ such that ${\displaystyle a_{n+1}\leq ca_{N}n^{-R}}$ for ${\displaystyle n\geq N}$; since ${\displaystyle R>1}$ this shows that ${\displaystyle \sum a_{n}}$ converges.

Special cases

All of the tests in De Morgan's hierarchy except Gauss's test can easily be seen as special cases of Kummer's test:^[4]

• For the ratio test, let ζ_n=1. Then:

${\displaystyle \rho _{\text{Kummer}}=\left({\frac {a_{n}}{a_{n+1}}}-1\right)=1/\rho _{\text{Ratio}}-1}$

• For Raabe's test, let ζ_n=n. Then:

${\displaystyle \rho _{\text{Kummer}}=\left(n{\frac {a_{n}}{a_{n+1}}}-(n+1)\right)=\rho _{\text{Raabe}}-1}$

• For Bertrand's test, let ζ_n=n ln(n). Then:

${\displaystyle \rho _{\text{Kummer}}=n\ln(n)\left({\frac {a_{n}}{a_{n+1}}}\right)-(n+1)\ln(n+1)}$

Using ${\displaystyle \ln(n+1)=\ln(n)+\ln(1+1/n)}$ and approximating ${\displaystyle \ln(1+1/n)\rightarrow 1/n}$ for large n, which is negligible compared to the other terms, ${\displaystyle \rho _{\text{Kummer}}}$ may be written:

${\displaystyle \rho _{\text{Kummer}}=n\ln(n)\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\ln(n)-1=\rho _{\text{Bertrand}}-1}$

• For Extended Bertrand's test, let ${\displaystyle \zeta _{n}=n\prod _{k=1}^{K}\ln _{(k)}(n).}$ From the Taylor series expansion for large ${\displaystyle n}$ we arrive at the approximation

${\displaystyle \ln _{(k)}(n+1)=\ln _{(k)}(n)+{\frac {1}{n\prod _{j=1}^{k-1}\ln _{(j)}(n)}}+O\left({\frac {1}{n^{2}}}\right),}$

where the empty product is assumed to be 1. Then,

${\displaystyle \rho _{\text{Kummer}}=n\prod _{k=1}^{K}\ln _{(k)}(n){\frac {a_{n}}{a_{n+1}}}-(n+1)\left[\prod _{k=1}^{K}\left(\ln _{(k)}(n)+{\frac {1}{n\prod _{j=1}^{k-1}\ln _{(j)}(n)}}\right)\right]+o(1)=n\prod _{k=1}^{K}\ln _{(k)}(n)\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\sum _{j=1}^{K}\prod _{k=1}^{j}\ln _{(K-k+1)}(n)-1+o(1).}$

Hence,

${\displaystyle \rho _{\text{Kummer}}=\rho _{\text{Extended Bertrand}}-1.}$

Note that for these four tests, the higher they are in the De Morgan hierarchy, the more slowly the ${\displaystyle 1/\zeta _{n}}$ series diverges.

Proof of Kummer's test

If ${\displaystyle \rho _{n}>0}$ then fix a positive number ${\displaystyle 0<\delta <\rho _{n}}$. There exists a natural number ${\displaystyle N}$ such that for every ${\displaystyle n>N,}$

${\displaystyle \delta \leq \zeta _{n}{\frac {a_{n}}{a_{n+1}}}-\zeta _{n+1}.}$

Since ${\displaystyle a_{n+1}>0}$, for every ${\displaystyle n>N,}$

${\displaystyle 0\leq \delta a_{n+1}\leq \zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}.}$

In particular ${\displaystyle \zeta _{n+1}a_{n+1}\leq \zeta _{n}a_{n}}$ for all ${\displaystyle n\geq N}$ which means that starting from the index ${\displaystyle N}$ the sequence ${\displaystyle \zeta _{n}a_{n}>0}$ is monotonically decreasing and positive which in particular implies that it is bounded below by 0. Therefore, the limit

${\displaystyle \lim _{n\to \infty }\zeta _{n}a_{n}=L}$ exists.

This implies that the positive telescoping series

${\displaystyle \sum _{n=1}^{\infty }\left(\zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}\right)}$ is convergent,

and since for all ${\displaystyle n>N,}$

${\displaystyle \delta a_{n+1}\leq \zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}}$

by the direct comparison test for positive series, the series ${\displaystyle \sum _{n=1}^{\infty }\delta a_{n+1}}$ is convergent.

On the other hand, if ${\displaystyle \rho <0}$, then there is an N such that ${\displaystyle \zeta _{n}a_{n}}$ is increasing for ${\displaystyle n>N}$. In particular, there exists an ${\displaystyle \epsilon >0}$ for which ${\displaystyle \zeta _{n}a_{n}>\epsilon }$ for all ${\displaystyle n>N}$, and so ${\displaystyle \sum _{n}a_{n}=\sum _{n}{\frac {a_{n}\zeta _{n}}{\zeta _{n}}}}$ diverges by comparison with ${\displaystyle \sum _{n}{\frac {\epsilon }{\zeta _{n}}}}$.