Question

help pleaseee...

how do i know its alternating series ?
number 11

Answer 1

if you see \((-1)^n\) it is for sure

Answer 2

or \((-1)^{n+1}\) in your case

Answer 3

okay so i plugged in numbers and im aussuming its decreasing

Answer 4

of course it is decreasing
you have a rational function where the degree of the numerator is smaller than the degree of the denominator

Answer 5

in some pre-calc class you might remember that such a function has a horizontal asymptote at \(y=0\) which is another way of saying the numbers get closer to zero

Answer 6

oh riight i remember

Answer 7

okay so how would i solve this problem because i know it starts like this

1. )bn
\[\frac{ n^2 }{ n^3+4} >0\]

Answer 8

we know it alternates right?

Answer 9

so all you need for convergence is that the terms go to zero

Answer 10

hmm okay but does this require derivatives ?

Answer 11

you could do that if you like
you could take the derivative say "hmmm it is always negative, so the function is decreasing and therefore the terms of the series are decreasing as well"

Answer 12

you could do a harder method, say show that \(|a_{n+1}|<|a_n|\) but that is a bunch of work

Answer 13

actually now that i look closer, you have to be a bit careful
did you take the derivative of \(\frac{x^2}{x^3+4}\)?

Answer 14

oh gawd lool is there an easier way do this D:

Answer 15

yeah the derivative is the easy way, then i will tell you an even easier way, but your math teacher may not accept it

Answer 16

hmm not yet but i get confuse with these things,,,

is this step necessarily ?
b n+1
this is where we plugin n+1 into the original equation ?

Answer 17

yes

Answer 18

oh i mean "yes, that is \(b_{n+1}\) "but yo don't necessarily need it

Answer 19

just take the derivative, what do you get?

Answer 20

if you are stuck let me know

Answer 21

okay so we would use qoutient rule


\[\frac{ (n^2)'(n^3+4)- (n^2)(n^3+4)' }{ (n^3+4)^2 }\]

Answer 22

jeez yes but it is not so bad as you wrote it

Answer 23

and although it doesn't matter, it should be \(x\) not \(n\)
\(n) are integers 1

Answer 24

\[\frac{2x(x^3+4)-3x^2(x^2)}{(x^2+4)^2}\]

Answer 25

\[\frac{ (2n)(n^3+4)-(n^2)(3n^2)}{ (n^6+8n^3+16) } = \frac{ 2n^4+8n-3n^4 }{ n^6+8n^3+16}\]

Answer 26

don't bother with multiplying the denominator

Answer 27

ok typo there

Answer 28

\[\frac{-x^4+8x}{(x^3+4)^2}\]

Answer 29

got that ?

Answer 30

oh riight so in the qoutient rule we multiply the bottom at all ?

Answer 31

now the denomiator is always positive, because it is a square
the numerator is NOT always negative,
however, it is always negative if \(x>2\) which means for \(n>2\) the sequence of terms is decreasnig

Answer 32

yeah you do get the denominator squared, i was just saying don't square it
it doesn't help your cause
easier to see that the denominator is always positive if you leave it as a square

Answer 33

oh okay question how did you get x>2 and n >2 ?

Answer 34

so if you want to make your math teacher happy, say
"consider \(f(x)=\frac{x^2}{x^2+4}\) whose derivative if \[f'(x)=-\frac{x(x^3-8)}{(x^3+4)^2}\] we see that for \(x>2\) we have \(f'(x)<0\) so for \(n>2\) the series is decreasing

Answer 35

how did i get \(x>2\)?

Answer 36

it should be clear that if \(x>2\) then \(x^3-8>0\) and also \(x(x^3-8)>0\) so \(-x(x^3-8)<0\)

Answer 37

this is really really too much work
the terms are all positive
they also go to zero because the numerator has degree less than the denominator
if the terms go to zero and they are all positive, then they must decrease by sheer obviousness

Answer 38

hmm i think the 2 is confusing me D:

Answer 39

If the larger series converges, then yours converges as well ...
(no need for that very left side in each line)

Answer 40

oh i think you got 2 by solving this right ? x^3-8 = 0 which is 2

Answer 41

yes

Answer 42

oh okay so then

Answer 43

is summery, if \(x>2\) you know \(f'(x)<0\) so if \(n>2\) the sequence of absolute values decreases to zero
since the series alternates that shows convergence of the series

Answer 44

*summary

Answer 45

ohh okay now it makes sense .but every alternate series does it always converge ?

Answer 46

not if the terms don’t go to zero!!

Answer 47

\[\sum(-1)^nn\] doesn't converge

Answer 48

for that matter, neither does \[\sum(-1)^n\]

Answer 49

ohhh but do we have to find the limit for this one?

Answer 50

which one?

Answer 51

i see one in your worksheet that does not converge
do you see it?

Answer 52

oh perhaps you are asking if you have to actually compute the sum
that is very very hard if not impossible
so no

Answer 53

oh the one we did right now do we have to find the limit like this