Kummer's test - Calculus, Apostol, 10.16 #15.
I want to prove the following:
Let $\{ a_n \}$ and $\{ b_n \}$ be two sequences with $a_n>0$ and $b_n>0$ for all $n \geq N$, and let $$c_n = b_n - \frac{a_{n+1}b_{n+1}}{a_n}$$
Then
- If there exists $r>0$ such that $0<r\leq c_n \text{ ; } \forall n\geq N$ then $\displaystyle\sum a_n$ converges.
- If $c_n\leq0$ for $n\geq N$ and if $\displaystyle\sum \dfrac{1}{b_n}$ diverges, then so does $\displaystyle\sum a_n$.
So far I know how to use and prove Cauchy's, D'Alambert's and the integral criterion, so I'd like a hint on either using those or a new idea. The book suggests that for $1$, I show that
$$\sum\limits_{k = N}^n {{a_k} \leq \frac{{{a_N}{b_N}}}{r}} $$
and for $2$, to prove that $\displaystyle\sum a_n$ dominates $\displaystyle\sum \dfrac{1}{b_n}$
I'd like to prove this with previous theory on series convergence (Cauchy's and/or D'Alambert's preferrably, comparison test) since it is what precedes the problem, and would appreaciate great HINTS rather than answers.
I can't seem to understand the inequalities. I mean, Cauchy's and D'Alambert's criteria reveal that the proof relies on the fact that a geometric series of with ratio $|r| < 1$ converges, but I can't understand the motivation in this proof.
Summing up the work.
For 1:
$$\eqalign{ & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{b_{N + 1}} \cr & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{b_{N + 2}} \cr & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} + \frac{{{a_{N + 3}}}}{{{a_N}}}{b_{N + 3}} \cr} $$
Induction over $n$ we get
$${b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}{c_{N + n}} + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}}$$
So
$$\eqalign{ & {b_N} \geqslant r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}} \right) + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}} \cr & {b_N} > r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}} \right) \cr & \frac{{{a_N}{b_N}}}{r} > {a_N} + {a_{N + 1}} + {a_{N + 2}} + \cdots + {a_{N + n}} = \sum\limits_{k = N}^n {{a_k}} \cr} $$
As desired.
Here’s a hint for (1):
$$\begin{align*}b_N &= c_N + \frac{a_{N+1}b_{N+1}}{a_N}\\ &=c_N+\frac{a_{N+1}}{a_N}\left(c_{N+1}+\frac{a_{N+2}b_{N+2}}{a_{N+1}}\right)\\ &=c_N+\frac{a_{N+1}}{a_N}\cdot c_{N+1}+\frac{a_{N+2}}{a_N}\cdot b_{N+2}\\ &=c_N+\frac{a_{N+1}}{a_N}\cdot c_{N+1}+\frac{a_{N+2}}{a_N}\cdot c_{N+2}+\frac{a_{N+3}}{a_N}\cdot b_{N+3}\\ &\ge r\left(1+\frac{a_{N+1}}{a_N}+\frac{a_{N+2}}{a_N}\right)+\frac{a_{N+3}}{a_N}\cdot b_{N+3}\tag{why?}\\ &=\frac{r}{a_N}(a_N+a_{N+1}+a_{N+2})+\frac{a_{N+3}}{a_N}\cdot b_{N+3}\\ &>\frac{r}{a_N}(a_N+a_{N+1}+a_{N+2})\;.\tag{why?} \end{align*}$$
Generalize to get an upper bound on the partial sums of $\sum a_n$.
Added: A hint for (2): if $$b_n\le\frac{a_{n+1}b_{n+1}}{a_n}\;,$$ then $$\frac{a_n}{1/b_n}=a_nb_n\le a_{n+1}b_{n+1}=\frac{a_{n+1}}{1/b_{n+1}}\;.$$
Since $a_{n} > 0$ and $c_{n}\ge r$, we have $$ a_{k}b_{k} - a_{k+1}b_{k+1} \ge r a_{k}. $$ Summing up from $k=N$ to $k=n$, $$ \sum_{k=N}^{n}\left(a_{k}b_{k}-a_{k+1}b_{k+1}\right) \ge r\sum_{k=N}^{n}a_{k}. $$ The LHS is a telescopic sum, yielding $$ \sum_{k=N}^{n}a_{k} \le \frac{a_{N}b_{N} - a_{n+1}b_{n+1}}{r}. $$ Since $a_n, b_n >0$, we get the desired upper bound: $$ \sum_{k=N}^{n}a_{k}\le\frac{a_{N}b_{N}}{r}. $$
The second inequality may be deduced by the same fashion.