How to prove the Bertrand convergence test that states:
If there exists such limit that $$\lim_{n\to\infty}((n(\frac{a_n}{a_{n+1}}-1)-1)\ln{n})=q$$ then the series $\sum^{+\infty}_{n=1}=a_n$, where $a_n>0, \forall n\in\mathbb{N}$ , converges if $q>1$ and diverges if $q<1$.
The problem is from the book Drugi, D., Collection of Problems in Mathematical Analysis (in Serbian), Naša Knjiga D.O,O., 2003, pp. 38, problem 16.
I have preference for considering rations $\frac{|a_{n+1}|}{|a_n|}$ instead of $\frac{|a_n|}{|a_{n+1}|}$. The former setting is in a way more consistent with how ratio test in presented in Calculus courses, the latter setting is used in the following Wikipedia article.
All the result will be presented in the former setting. It is not too difficult to make the appropriate modifications to express the results the way if Wikipedia.
Bertrand's and Raabe's tests can be viewed as particular instances of Kummer's test:
Kummer: Let $\sum_na_n$ be a series with terms in $\mathbb{C}\setminus\{0\}$ and $b_n$ a sequence of positive numbers.
- (a) If there is $r>0$ and $N\in\mathbb{N}$ such that \begin{align} b_n-b_{n+1}\frac{|a_{n+1}|}{|a_n|}\geq r,\qquad n\geq N\tag{1}\label{one} \end{align} then $\sum_na_n$ is absolutely convergent.
- (b) If there is $N\in\mathbb{N}$ such that \begin{align} b_n-b_{n+1}\frac{|a_{n+1}|}{|a_n|}\leq 0,\qquad n\geq N\tag{2}\label{two} \end{align} and $\sum_n\frac1{b_n}$ diverges, then $\sum_n|a_n|$ diverges.
Proof: (a) For all $n\geq N$, $$b_n|a_n|-b_{n+1}|a_{n+1}|\geq r|a_n|\geq0$$ Adding terms from between $N$ and $N$ yields $$b_N|a_N|\geq b_N|a_N|-b_{n+1}|a_{n+1}|\geq r\sum^n_{k=N}|a_k|$$ This shows that $\sum^n_{k=1}|a_k|$ is a bounded monotone nondecreasing sequence.
(b) It follows that formal $n\neq N$ $$\frac{|a_n|}{1/b_n}\leq \frac{|a_{n+1}|}{1/b_{n+1}}$$ Thus, the sequence $n\mapsto \frac{|a_n|}{1/b_n}$ is monotone nondecreasing and bounded below; hence $$c\frac{1}{b_n}\leq |a_n|,\qquad n\geq N$$ for some $c>0$. The conclusion follows from comparison.
The sequence $b_n=(n-1)\log(n-1)$ corresponds to Bertrand's test.
Bertrand: Suppose $|a_n|>0$ for all $n\geq N$
- (c) If $\frac{|a_{n+1}|}{|a_n|}\leq 1-\frac1n-\frac{q_n}{n\log n}$ such that $q_n\geq q>1$ for all $n\geq N$, then $\sum_na_n$ converges absolutely.
- (d) If $\frac{|a_{n+1}|}{|a_n|}\geq 1-\frac1n-\frac{q_n}{n\log n}$ such that $q_n\leq q<1$ for all $n\geq N$, then $\sum_n|a_n|$ diverges.
In many instances, one has $$\lim_n n\log n\big(1 -\tfrac{|a_{n+1}|}{|a_n|}\big)-\log n =q $$ exists. If $q>1$, then $\sum_n|a_n|$ converges while if $q<1$, $\sum_n|a_n|$ diverges. When $q=1$, the test is inconclusive.
Proof: (c) implies that if $n\geq N$ $$(n-1)\log(n-1) -n\log(n)\frac{|a_{n+1}|}{|a_n|}\geq q_n+\log\big(1-\tfrac1n\big)^{n-1}\geq q+\log\big(1-\tfrac1n\big)^{n-1}$$ As $\log\big(1-\tfrac1n\big)^{n-1}\xrightarrow{n\rightarrow\infty}-1$, there is $N'\geq N$ such that $\log\big(1-\tfrac1n\big)^{n-1}>-1-\frac{q-1}{2}$ for $n\geq N'$. Hence, for $n\geq N'$ $$(n-1)\log(n-1) -n\log(n)\frac{|a_{n+1}|}{|a_n|}\geq q-1-\frac{q-1}{2}=\frac{q-1}{2}>0$$
(d) implies that if $n\geq N$ $$(n-1)\log(n-1) -n\log(n)\frac{|a_{n+1}|}{|a_n|}\leq q_n+\log\big(1-\tfrac1n\big)^{n-1}\leq q+\log\big(1-\tfrac1n\big)^{n-1}$$ As $\log\big(1-\tfrac1n\big)^{n-1}\xrightarrow{n\rightarrow\infty}-1$, there is $N'\geq N$ such that $\log\big(1-\tfrac1n\big)^{n-1}<-1+\frac{1-q}{2}$ for $n\geq N'$. Hence, for $n\geq N'$ $$(n-1)\log(n-1) -n\log(n)\frac{|a_{n+1}|}{|a_n|}\leq q+\log\big(1-\tfrac1n\big)^{n-1}<-\frac{1-q}{2}<0$$
The sequence $b_n=n-1$ corresponds to Raabe's test:
Raabe: Suppose$|a_n|>0$ for all $n\geq N$.
- (e) If $\frac{|a_{n+1}|}{|a_n|}\leq 1-\frac{p}{n}$ for some $p>1$, $\sum_na_n$ converges absolutely.
- (f) If $\frac{|a_{n+1}|}{|a_n|}\geq 1-\frac{p}{n}$ for some $p\leq1$, $\sum_n|a_n|$ diverges.
In many instances, one has $$q=\lim_n n\big(1-\tfrac{|a_{n+1}|}{|a_n|}\big)$$ exists. If $q>1$, $\sum_n|a_n|$ converges while if $q<1$, $\sum_n|a_n|$ diverges. When $q=1$, the test is inconclusive.
Proof: Let $b_n=n-1$.
(e) implies that $$ b_n- b_{n+1}\frac{|a_{n+1}|}{|a_n|}\geq p-1>0$$
(f) implies that $$b_n- b_{n+1}\frac{|a_{n+1}|}{|a_n|}\leq p-1\leq 0$$
Comment: Notice that the limit version of Bertrand's theorem can be seen as an improvement to Raabe's test when the latter fails:
$$\lim_n n\log n\Big(1-\tfrac{|a_{n+1}|}{|a_n|}\Big)-\log n=\lim_n\Big(n\big(1-\tfrac{|a_{n+1}|}{|a_n|}\big)-1\Big)\log n$$
