Question

Hi everyone! :o) I need a little help! Diverges or Converges via Ratio Test? Sigma n=7 to infinity of (n+7)!/(2n-4)!...I posted a pic of my work as 23.bmp...can someone look at my work and discuss with me if I did this correctly? Thanks! :o)

Answer 1

Well done
\[\Huge \checkmark \]

Answer 2

hi Spacelimbus! Thanks...I appreciate the check!...I do have a small question though...

Answer 3

did I forget absolute value signs at all?

Answer 4

Why do you consider absolute values? Because of absolute diverges or converges?

Answer 5

like towards the end when I have (2n-4) on top...should I have changed that to (2n+4) and then (2n+3) etc...?

Answer 6

I mean, if that is true, everything would still cancel and the limit would go to inifnity, but I was just worried I didn't do it perfectly even though the answer may be the same?

Answer 7

No you don't have to do that, keeping the signs as they appear in your work is fine. What is important are the powers of the resulting polynomial. Also consider that:
\[\Large |(2n-4)|=-2n+4 \]
So the your limit approaches negative infinity, it would still result to be positive infinity. But it's non important for that, because convergence is just within the radius 1,

\[\Large |r|\leq 1 \]

Answer 8

ahhh...ok...I see...great explanation...but just to add another dumb question...would it have been WRONG if I did what I suggested in turning the 2n-4 into 2n+4 and 2n+3 etc? I mean, If for some uncontrollable reason I just felt it necessary to do extra work, would it have been technically wrong? :o)

Answer 9

I'm just trying to cememnt down my logic is all!

Answer 10

Oh I see now where and why you struggle I guess. I believe I have then commented a bit wrong on the given problem. Remember that n factorial is defined as:
\[\Large n!=n(n-1)(n-2)(n-3) \cdot ... \]
Similarly you have:
\[\Large (2n-3)!=(2n-3)(2n-4)(2n-5) \]

Answer 11

oh you are correct...I actually wrote the problem incorrectly! yikes...i should have written it (2n-4)(2n-5)(2n-6) etc... I mean they still cancel but technically wrong it looks like!

Answer 12

So you might want to review that part of your proof:
\[\Large \frac{(n+8)!}{(2n-3)!}\cdot \frac{(2n-4)!}{(n+7)!} \]
I'd much rather write that out as:
\[\Large \frac{(n+8)(n+7)! \cdot (2n-4)!}{(2n-3)(2n-4)! (n+7)!} \]

Answer 13

\[\Large \frac{(n+8)}{(2n-3)}= \frac{1}{2} \]
Converges. I am sorry.

Answer 14

c'mon spacelimbas

Answer 15

yeah...if you don't mind, I would like to repost the corrections...would you take a look at them for me in a couple minutes Spacelimbus?

Answer 16

where did I go wrong @cblrtopas ?

Answer 17

and sure @SinginDaCalc2Blues

Answer 18

You're applying the Ratio test incorrectly; 2n-4 --> 2(n+1)-4=2n-2 not 2n-3

Answer 19

Yes you're right @cblrtopas , thanks for pointing that out. It's a wrong substitution in the first place to start with.

Answer 20

Geez...so I did it wrong for a second time?

Answer 21

well for the denominator, it should more turn out to be like
\[\Large \frac{(n+8)!(2n-4)!}{(2n-2)!(n+7)!} \]

Answer 22

ok...so what you are saying then if I understand correctly...when I apply the factorial notation written out, whatever it is, like (2n-129), I would single out the "n" and add a one to it?

Answer 23

2(n+1)-129 ?

Answer 24

i thought factorial was reducing the quantity by 1, not adding a 1 ???

Answer 25

That is correct @SinginDaCalc2Blues, what @cblrtopas did point out is that the substitution at the beginning was wrong. You replace n by n+1

So (n+7) becomes (n+1+7)=(n+8)
Same goes for (2n-4) it becomes (2(n+1)-4)=(2n-2)

Answer 26

i am disagreeing with that cblrt guy...i already added the n+1 to my problem at the beginning!

Answer 27

you disagreeing with da trufffff

Answer 28

he's right, you multiply, you don't add it.

Answer 29

just kidding you got it Calcblues

Answer 30

yeah limbus...i did that at the beginning of the problem...check out my work...i dont see where i did anything wrong

Answer 31

CalcBlues you have to make a substitution; you replace n with n+1 wherever it exists in the series. This is akin to multiplication try it again --- very important that you get this part down

Answer 32

Yeh some confusion emerged here I agree ^^

But the two is a scalar multiplication, check above again, you do not add (n+1) you replace (n) by (n+1) so if there is a scalar in front of it, like 2, you multiply the entire expression by 2, not add it.

Answer 33

okay...i understand what you are saying..i do...just hear me out...

Answer 34

Look the answer is still correct thanks to L'Hopital's rule

Answer 35

nevermind...i just saw it!

Answer 36

whatever's after the 2n is obliterated anyway so no sweat

Answer 37

i must distrubute the 2

Answer 38

yeh it turns out to be right by a miracle ^^

Answer 39

ok guys...i'm gonna work it one more time...can you check it for me in a couple minutes again? :o)

Answer 40

But @cblrtopas was right to highlight that, it's a minor mistake, but it would still cost valuable points during an exam.

Answer 41

almost done...

Answer 42

it's the same as your other answer you just change one term in the denominator and you're done.

Answer 43

Not quite the same as for the final result, it now converges to zero.

Answer 44

you serious?

Answer 45

let me look at this

Answer 46

There it is...converges to zero...whew...please tell me this is correct!!!! please!!! :0)

Answer 47

now everything looks right to me (-: Yes.

Answer 48

does cblrstuff agree?

Answer 49

It looks correct

Answer 50

I hate to think about what would have happened to me on my exam if we hadn't noticed this error! omg

Answer 51

The good thing is that we most likely did every possible mistake in that journey, so we can still say in conclusion that it was a fabulous exercise *grins*.

Answer 52

you would be beaten with a yardstick.

Answer 53

oh wait!!!...i see another problem...it's not correct!

Answer 54

girl just move on to the next problem.

Answer 55

j/k...muahahaha...it isn't possible to see another problem...we already made every error! haha

Answer 56

can I give 2 medals or just one?

Answer 57

^^

Answer 58

you got me (no you didn't)

Answer 59

take it easy lame-o

Answer 60

darn...just one medal...hey cblrstuff&stuff...i owe you a medal!!! :o)

Answer 61

well at least I learned a ton about the ratio test and factorials! yay! :o) thanks guys soooo much.. i really appreciate all your help!...seriously! :o)