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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….

$a_{n+1}-a_n=\frac{n^2+2n-n^2-2n-1}{n(n+1)}=\frac{-1}{n(n+1)}<0$

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 $(-1)^n$?

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

$a_{n+1}<a_n$

Depending on which is more convenient you get

$\frac{a_{n+1}}{a_n}<1$

and

$a_{n+1}-a_n<0$

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 $0$

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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.

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    $\begingroup$ 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
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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 $a_n > a_{n+1}$.

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