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Mathematics 65 Online
OpenStudy (astrophysics):

Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?

OpenStudy (astrophysics):

\[a_n = \frac{ 2n-3 }{ 3n+4 }\] @ganeshie8

OpenStudy (astrophysics):

I don't specifically understand these problems, as in the (n+1) business and what not, so could you help guide me through this?

OpenStudy (astrophysics):

\[a_n+1 = \frac{ 2(n+1)-3 }{ 3(n+1)+4 } \] right?

ganeshie8 (ganeshie8):

yes, you want to show \(\large a_n < a_{n+1}\) is it ?

OpenStudy (astrophysics):

Yeah something of that sort

OpenStudy (astrophysics):

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?

OpenStudy (astrophysics):

I'll brb ganeshie, I'm sorry.

OpenStudy (ikram002p):

well a_(n+1)- a_n to know if this value >0 then increasing <0 decreasing -_- simply like this

ganeshie8 (ganeshie8):

Ahh nice trick :)

OpenStudy (ikram002p):

no difference lol btw saying this or saying a_n+1>a_n hmmm

ganeshie8 (ganeshie8):

for this particular sequence, taking ratio or finding derivative gives messy things compared to taking the difference

jagr2713 (jagr2713):

@ganeshie8 can you help me?

ganeshie8 (ganeshie8):

\[\large \begin{align} a_{n+1}-a_n &= \dfrac{2(n+1)-3}{3(n+1)+4} - \dfrac{2n-3}{3n+4}\\~\\& = \dfrac{17}{(3n+7)(3n+4)}\\~\\& \gt 0\end{align}\]

OpenStudy (ikram002p):

haha yeah right:P and its bounded , right ?

OpenStudy (astrophysics):

Yes, it's bounded, just doing it you get 2/3

OpenStudy (astrophysics):

Wait hold on

OpenStudy (astrophysics):

Sequence is bounded below by the first term - 1/7?

ganeshie8 (ganeshie8):

taking the limit \(\large n\to \infty\) should do for boundedness i think

OpenStudy (ikram002p):

sry inf of -3/4 :P typo and sup of 2/3

OpenStudy (astrophysics):

Boo yeah!

OpenStudy (ikram002p):

first term ? idk what u ppl use a_0 or a_1 :D :P :P

jagr2713 (jagr2713):

@ganeshie8

ganeshie8 (ganeshie8):

check ur msg

OpenStudy (astrophysics):

Thanks guys :)

OpenStudy (astrophysics):

But, so if I do a_n+1 only I would have to take the derivative and stuff right?

OpenStudy (ikram002p):

why O.O what test u wanna do ?

ganeshie8 (ganeshie8):

f'(x) > 0 ==> increasing f'(x) < 0 ==> decreasing

ganeshie8 (ganeshie8):

sequence follows the corresponding function right ?

OpenStudy (ikram002p):

nope

OpenStudy (ikram002p):

sequence are discreat

ganeshie8 (ganeshie8):

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