Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?
\[a_n = \frac{ 2n-3 }{ 3n+4 }\] @ganeshie8
I don't specifically understand these problems, as in the (n+1) business and what not, so could you help guide me through this?
\[a_n+1 = \frac{ 2(n+1)-3 }{ 3(n+1)+4 } \] right?
yes, you want to show \(\large a_n < a_{n+1}\) is it ?
Yeah something of that sort
But, in notes it says a problem as such should be \[a_n+1-a_n \]I don't see why it's - a_n?
I'll brb ganeshie, I'm sorry.
well a_(n+1)- a_n to know if this value >0 then increasing <0 decreasing -_- simply like this
Ahh nice trick :)
no difference lol btw saying this or saying a_n+1>a_n hmmm
for this particular sequence, taking ratio or finding derivative gives messy things compared to taking the difference
@ganeshie8 can you help me?
\[\large \begin{align} a_{n+1}-a_n &= \dfrac{2(n+1)-3}{3(n+1)+4} - \dfrac{2n-3}{3n+4}\\~\\& = \dfrac{17}{(3n+7)(3n+4)}\\~\\& \gt 0\end{align}\]
haha yeah right:P and its bounded , right ?
Yes, it's bounded, just doing it you get 2/3
Wait hold on
Sequence is bounded below by the first term - 1/7?
taking the limit \(\large n\to \infty\) should do for boundedness i think
sry inf of -3/4 :P typo and sup of 2/3
Boo yeah!
first term ? idk what u ppl use a_0 or a_1 :D :P :P
@ganeshie8
check ur msg
Thanks guys :)
But, so if I do a_n+1 only I would have to take the derivative and stuff right?
why O.O what test u wanna do ?
f'(x) > 0 ==> increasing f'(x) < 0 ==> decreasing
sequence follows the corresponding function right ?
nope
sequence are discreat
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