Sequences~
Determine whether \(\Large a_{n}={\frac{3+5n^2}{n+n^2}}\) diverges or converges. Also determine the sequence's tendency and find it's bounds
@ganeshie8 ?
@iambatman @ikram002p
@thomaster
@ParthKohli
so what test did u try ?
@Ashleyisakitty
I have had no practice in this area so sorry if Idk what you're talking about.. but I did the divergence test.
do u think its diverge or converge ? (are you familier with comparasion test or ratio test ?)
this is just a sequence right ? @ikram002p
yes mm so ?
It diverges since the limit was not 0.. And I know the ratio test
looks we're mixing sequence and series
ohhh , ok i got ur point but i was thinking if the series was converge then the seqence is converge as well
but finding the limit as a_n goes to infinity would be better , sry my bad lol
yes !
So based on the sequence test.. since it doesn't go to \(\infty\) it converges?
yes \(\large \lim \limits_{n\to \infty }a_{n}=\lim \limits_{n\to \infty }{\dfrac{3+5n^2}{n+n^2}} = \dfrac{5}{1}\)
since the limit exists, the sequence converges.
How would I find it's tendencies and bounds?
\[\large A \le \dfrac{3+5n^2}{n+n^2} \le B\]
I'm also not sure what they mean by `tendencies`, but we can try finding the bounds ^ @ikram002p help
we already have the upper bound \[\large A \le \dfrac{3+5n^2}{n+n^2} \lt 5\]
@dan815
Luigi what does it mean tendency?
tendencies is the limit when n goes to infinity bounded for sure 0<a_n<=5 mmm u can found another lower bound
the mean of the sequence?
oh ok
@ganeshie8 so what do you think ? n should be integer right ? no need to minimize since its monotone \(a_1=4\) and upper bound 5
so 4<=a_n<5 right ?
yes its a monotone sequence for n > 3 check a_2 once
3.8
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