Let $\{a_n\}, \{b_n\}$ be positive sequences. Let $c_n= b_n-\dfrac {b_{n+1}a_{n+1}} {a_n}$. Prove that :

If there is a positive constant $r$ such that $c_n \geq r >0 ~\forall~n\in \mathbb N$, then $\sum_{n=1}^\infty a_n $ converges and ..... If $c_n \leq 0~\forall~n\in \mathbb N$ and if $\sum \dfrac {1}{b_n}$ diverges, so does $\sum a_n$.

Attempt: I understand the above problem statement as : For every $r >0, ~\exists~ N \in \mathbb N$, such that $c_n \geq r >0$ whenever $n \geq N$.

$\implies $ the series $c_n$ is unbounded and $\lim_{n \rightarrow \infty} c_n = \infty $

$\implies \lim_{n \rightarrow \infty} b_n - \dfrac {b_{n+1}a_{n+1}} {a_n} = \infty$

$\implies $ for a sufficiently large $n, b_n >> \dfrac {b_{n+1}a_{n+1}} {a_n}$

$\implies \lim_{n \rightarrow \infty} \dfrac {b_{n+1}~a_{n+1}} {b_n~a_n}=0$

How should I proceed from here?

Thank you for your help in this regard.

  • $\begingroup$ Is there a chance you meant to write in your second line "...then $\;\sum a_n\;$ converges if $\;\sum\frac1{b_n}\;$ converges? $\endgroup$ – Timbuc Jan 24 '15 at 12:25
  • $\begingroup$ uhm no actually.. it's the same. $\endgroup$ – MathMan Jan 24 '15 at 12:31
  • $\begingroup$ Ok, I see. Thank you. $\endgroup$ – Timbuc Jan 24 '15 at 12:36

$c_n= b_n-\dfrac {b_{n+1}a_{n+1}} {a_n} \Rightarrow a_nc_n = b_{n}a_{n}-b_{n+1}a_{n+1} \Rightarrow \sum_{n=1}^m (a_nc_n) = b_1a_1-b_{m+1}a_{m+1}$

From $c_n \ge 0$ , find $b_na_n \ge b_{m+1}a_{m+1}$ . Thus suppose $M = b_1a_1 \Rightarrow \forall n \in \mathbb N \ , M \ge b_na_n$

Then $\forall m \in \mathbb N ,\ \sum_{n=1}^m(ra_n) \le \sum_{n=1}^m (a_nc_n) = b_1a_1-b_{m+1}a_{m+1} \le M$

Hence $\sum_{n=1}^{\infty} a_n = \lim_{m \rightarrow \infty} \sum_{n=1}^m(a_n) \le \frac Mr$ is boundary , so converges since $a_n >0$.

$ \ $

From $c_n \le 0\Rightarrow b_na_n \le b_{m+1}a_{m+1}$ , still suppose $M = b_1a_1$ , get $M\le b_na_n \Rightarrow \frac{M}{b_n}\le a_n$

So $\sum_{n=1}^{\infty} a_n \ge \sum_{n=1}^{\infty} (\frac{M}{b_n}) = \infty $ , which means $\sum_{n=1}^{\infty} a_n$ diverges.

  • $\begingroup$ Thank you for the wonderful answer :) ! $\endgroup$ – MathMan Jan 24 '15 at 13:32

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