$a_n$ and $b_n$ are sequences of non-negative terms

$\dfrac{a_{n+1}}{a_n} \leq \dfrac{b_{n+1}}{b_n}$ for all $n$

If $\sum\limits_{n=1}^\infty b_n$ converges, prove that $\sum\limits_{n=1}^\infty a_n$ converges

If $\sum\limits_{n=1}^\infty a_n$ diverges, prove that $\sum\limits_{n=1}^\infty b_n$ diverges

I was thinking of trying the ratio test. For the 1st part, $\sum b_n$ converges, which implies that $\lim\limits_{n\to\infty}\left(\dfrac{b_{n+1}}{b_n}\right)$ is less than or equal to $1$. If it is less than $1$, than the same limit for $\frac{a_{n+1}}{a_n}$ is less than one, which implies $\sum\limits_{n=1}^\infty a_n$ converges. But I am not sure what to do about if the limit equals $1$.

Thanks for the help

  • $\begingroup$ You should take the time to format your post properly using MathJax. As it is written, it is unclear which of the following is intended: $a_n+\frac{1}{a_n}, \frac{a_n + 1}{a_n}, \frac{a_{n+1}}{a_n}$. (It is very likely you mean the third, but this should be made clear by you) $\endgroup$ – JMoravitz Aug 19 '16 at 19:23
  • $\begingroup$ As an aside, $\sum_{n=1}^\infty b_n$ converging does not imply that $\lim\limits_{n\to\infty} \frac{b_{n+1}}{b_n}$ exists. IF it exists, then as you say it could not be that the limit is greater than one, but it could be that $b_n = \frac{1}{2+(-1)^n}b_{n-1}$ and the sequence $\frac{b_{n+1}}{b_n}$ alternates between $\frac{1}{3}$ and $1$, not having a limit. Other counterexamples exist as well. Similarly for $a_n$ $\endgroup$ – JMoravitz Aug 19 '16 at 19:33
  • $\begingroup$ hmm interesting---what is the right way to approach this problem then? $\endgroup$ – Joseph Aug 19 '16 at 19:43

You have the ratio test backwards. It says if $b_{n+1}/b_n \to L <1,$ then $\sum b_n$ converges. It does not say that if $\sum b_n$ converges, then something or other is true about the sequence $b_{n+1}/b_n.$

Note that we need to assume all terms are positive, otherwise there is a problem with the ratios.


$$a_n = \frac{a_n}{a_{n-1}}\frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_2}{a_1}a_1.$$

The same is of course true for the $b_n$'s.


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