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
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.