From my notes from lecture:

${a_n}$ is increasing if $a_{n+1}-a_n>0$ and decreasing if $a_{n+1}-a_n<0$

When my professor worked through a problem, he explicitly left $a_{n+1}$ and $a_n$ on the same side of the inequality (as seen above).

So, for example, to show that ${a_n}={}\frac{n+1}{n}$ is decreasing….


1) How can I know that $\frac{-1}{n(n+1)}<0$?

2) Is there are reason for leaving both a_n and a_{n+1} on the same side? Does it not make more sense to use $a_{n+1}>a_n$ and $a_{n+1}<a_n$?

EDIT: As a follow up question to my answer below, what about when $a_n$ includes $(-1)^n$?

  • $\begingroup$ Is it always assumed that $n>1$? $\endgroup$ – rainier May 10 '18 at 3:34

1)You have a negative (and non-zero) numerator and an always positive denominator $(n\in\mathbb{N})$ so $-\frac{1}{n(n+1)}$ is always less than $0$

2)It's useful to use the definition


Depending on which is more convenient you get




Similarly for increasing sequences

3) As for your follow up question, think about how an alternating sequence can only converge if the non-alternating part converges to $0$


Oh my...I just realized that $a_n$ is only for $n>1$. Therefore, $\frac{-1}{n(n+1)}$ has to be less than one.

  • 1
    $\begingroup$ That doesn't matter. What matters is if it is positive or negative. $\endgroup$ – marty cohen May 10 '18 at 4:09
  • $\begingroup$ @martycohen But the sign of $\frac{-1}{n(n+1)}$ depends on the sign on $n$. Since $n>1$, $\frac{-1}{n(n+1)}$ will always be negative. $\endgroup$ – rainier May 10 '18 at 13:17

You should not get hung up about what side the $a_n$'s are on. It matters what the sign is saying. That is for the case of monotone decreasing, $a_{n+1} < a_n$ is just the same as $a_n > a_{n+1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.