Math help...

[AZReDWiNG]

We're working with the ratio test to find if a series is convergent or divergent.

Here's the problem.

Series:

2 + (4/2^2) + (8/3^2) + (16/4^2)...

So I found that An = ((2^n)/(n^2)) and A(n+1) = ((2^[n+1])/([n+1]^2))

Lim(n-->∞) {[2^(n+1)]/[(n+1)^2]}/[(2^n)/(n^2)]

Lim(n-->∞) {[(2^n)*2]/(n^2 + 2n + 1)}/[(2^n)/(n^2)]

Lim(n-->∞) {[(2^n)*2]/(n^2 + 2n + 1)} * [(n^2)/2^n)]

Lim(n-->∞) 2/(2n+1) = 0 ---> series converges, r < 1

The book lists the answer as divergent. Any idea why?

EDIT: Figured it out. I divided n^2 when mathematically impossible. Next problem i still don't know.

Another problem: I can't find An to the following series:

1 + 3/(1*2*3) + 5/(1*2*3*4*5) +...

I can see the pattern but I don't know how to write it in terms of n.. any ideas?

EDIT: Jeez, I'm an idiot. Someone check to see if An = [n+(n-1)]/[n+(n-1)]!

And delete this thread.

[GeorgiaMiddie2]

the series acts like (1/n), which is a harmonic series, which is divergent. that's whay it is divergent. you're thinking of a p-series, which is (1/x^n). but this is not a p-series, so, p>1 does not work here.

[Aussie_kid]

what year/age of maths is that?

[brlacrosse50]

yeah what math is this? im totally lost

[AZReDWiNG]

what year/age of maths is that?

Pre-calc, though the concept of limit is a Calc 1 thing.

[AZReDWiNG]

Who wants to figure out this series?

1/2 - 3/8 + 9/32 ...

I got the denominator as 2^(2n-1) but I'm stuck on the numerator..

[GeorgiaMiddie2]

it's (-3)^n when n starts at 0

[funnyboy]

I AM CONFUSED

[AZReDWiNG]

it's (-3)^n when n starts at 0

But n starts at 1...

[goalieskcickay]

THANKS FOR THAT!!!

[AZReDWiNG]

THANKS FOR THAT!!!

Uh, what???

[AXL]

But n starts at 1...
if your teacher wants "n" to start at one, then simply change GeorgiaMiddie2's answer to (-3)^(n-1). :dummy: That way your first, and second(and blah blah) terms are equal to georgiamiddie2's first and second(and blah blah) terms, but you start at one instead of zero.

[AXL]

Who wants to figure out this series?

1/2 - 3/8 + 9/32 ...

I got the denominator as 2^(2n-1) but I'm stuck on the numerator..
Isnt this a simple geometric series? The multiplicator(whats it called, geometric mean?) is negative .75 or (-3/4). Once you know that dont you just say; (1/2)="n"sub1(in other words, "n" initial), and "n"subx=("n"sub1)(-.75)^(x-1)
In the above expression, x equals all integers greater or equal to one. x =1 would of course represent the first term in the series, also reffered to as nsub1, which equals (1/2).


I dont see why you need a numerator and a denominator. If im off my game someone let me know, its been a while since i studied limits.

By the way my quotations around n dont mean anything, its just to eliminate confusion since i couldnt actually write a little one or x below n.

[chsFIRElax15]

i'm just now learning arithmetic sequences and series.

[AZReDWiNG]

Isnt this a simple geometric series? The multiplicator(whats it called, geometric mean?) is negative .75 or (-3/4). Once you know that dont you just say; (1/2)="n"sub1(in other words, "n" initial), and "n"subx=("n"sub1)(-.75)^(x-1)
In the above expression, x equals all integers greater or equal to one. x =1 would of course represent the first term in the series, also reffered to as nsub1, which equals (1/2).


I dont see why you need a numerator and a denominator. If im off my game someone let me know, its been a while since i studied limits.

By the way my quotations around n dont mean anything, its just to eliminate confusion since i couldnt actually write a little one or x below n.

It is geometric, but we're supposed to apply the ratio test which is A(n-1)/An .

[OutBurst]

I am on vacation right now so don't start asking me anything about Pre-Calc until tomorrow.

[AXL]

It is geometric, but we're supposed to apply the ratio test which is A(n-1)/An .
Im interested, what does A represent? And what is the purpose of this ratio test? Does it simplify things somehow? Or does it simply describe the expression in a different manner(in which case it must be simple to convert my equation to "ratio test" form)?

[AZReDWiNG]

Im interested, what does A represent? And what is the purpose of this ratio test? Does it simplify things somehow? Or does it simply describe the expression in a different manner(in which case it must be simple to convert my equation to "ratio test" form)?

An is the nth term of any sequence. A just signifies term number ___. So A(1) means the first term of a sequence, An means the nth term of a series, and A(n+1) means the nth + 1 term of a series.

The ratio test is used to test if a series converges (has a limit) or diverges (has no limit) for a series that is neither arithmetic nor geometric.

[franks2089]

I feel really stupid after reading those equations.

[AXL]

An is the nth term of any sequence. A just signifies term number ___. So A(1) means the first term of a sequence, An means the nth term of a series, and A(n+1) means the nth + 1 term of a series.

The ratio test is used to test if a series converges (has a limit) or diverges (has no limit) for a series that is neither arithmetic nor geometric.
oh, so my "n"subx is the same as your An.

"we're supposed to apply the ratio test which is A(n-1)/An ." How does this tell you if the sequence is convergent or divergent? So lets say i took the second term, -3/8. I put this into the ratio test and get (1/2)/(-3/8). Where does that leave me?

[AZReDWiNG]

oh, so my "n"subx is the same as your An.

"we're supposed to apply the ratio test which is A(n-1)/An ." How does this tell you if the sequence is convergent or divergent? So lets say i took the second term, -3/8. I put this into the ratio test and get (1/2)/(-3/8). Where does that leave me?

Err, I meant A(n+1)/An

If the limit of the resultant is less than 1, the series is convergent. If the limit of the resultant is greater than 1, it's divergent. This is why simply dividing A(2) and A(1) doesn't work; you've got to find the general term for n and n+1.

[kryptic]

just wait til you guys get to Calculus AB, you don't really see these again, I mean I forgot how to do them...

[Lacrossedude24]

Wow, the first place you think of going to ask for help with math is the forums? Try a texbook? A fellow student? A parent? Sibling?

[AZReDWiNG]

Wow, the first place you think of going to ask for help with math is the forums? Try a texbook? A fellow student? A parent? Sibling?

Hey, GeorgiaMiddie2 and AXL figured it out, so...

[Lacrossedude24]

True, did they get back to you in time for when you had to have the problems done?

[AZReDWiNG]

True, did they get back to you in time for when you had to have the problems done?

Yup.

I've got the problem written on my paper, I just have to simplify it.

[GeorgiaMiddie2]

just wait til you guys get to Calculus AB, you don't really see these again, I mean I forgot how to do them...

Oh, but when you get to Calc 2 in college, you see em again... i just took a test on this junk, but the series and sequences were about 10 times more complicated.

[kryptic]

Yeah, that's the bad part..

[AXL]

Err, I meant A(n+1)/An

If the limit of the resultant is less than 1, the series is convergent. If the limit of the resultant is greater than 1, it's divergent. This is why simply dividing A(2) and A(1) doesn't work; you've got to find the general term for n and n+1. oh i understand now, thanks for taking the time to explain it. You did so quite a bit better then i would have expected from a pre-calc student to be honest.

[AZReDWiNG]

oh i understand now, thanks for taking the time to explain it. You did so quite a bit better then i would have expected from a pre-calc student to be honest.

I aim to please :D

Math, among other things, tends to be my forte..