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Null sequence question

I'm stuck on an annoyingly easy question about null sequences. :ashamed2:

Use the list of basic null sequences and the various rules to prove the following sequence is null:

{ (n^10 x 10^n) / n! }

basic null sequences are:
{1/n^p} p>0
{c^n} |c|<1
{n^p x c^n} p>0, |c|<1
{c^n/n!}
{n^p/n!} p>0

and the various rules are sum rule, product rule, multiple rule.

thanks!! :ahee:
(edited 12 years ago)
Reply 1
What about focusing on one of the sequences in this list, and using a bit of common sense about n10n^{10} and 10n10^n?
Reply 2
split off half the numerator, prove the remainder is null and use the multiple rule to show any multiple of the bit you cut off by a null sequence is null?

*crosses fingers*
Reply 3
Original post by gff
What about focusing on one of the sequences in this list, and using a bit of common sense about n10n^{10} and 10n10^n?


I've been trying to do that, im not quite sure what you mean about common sense about n^10 and 10^n?
Reply 4
Original post by sputum
split off half the numerator, prove the remainder is null and use the multiple rule to show any multiple of the bit you cut off by a null sequence is null?

*crosses fingers*


I thought about that but according to the multiple rule, the multiplier has to be a real number (i think?) and it wouldnt be multiplying it by a real number, it would be either n^10 or 10^n

or am i being silly
Reply 5
What are the "various rules" here? [I'm treating this as an exercise in "proof using only these facts", in which case we need to know what facts we can assume].
Reply 6
Original post by DFranklin
What are the "various rules" here? [I'm treating this as an exercise in "proof using only these facts", in which case we need to know what facts we can assume].


If {a} and {b} are null then
{a+b} is null - sum rule
{Ka} is null, K belongs to all real numbers - multiple rule
{ab} is null - product rule
Reply 7
Original post by mootasaurous
I've been trying to do that, im not quite sure what you mean about common sense about n^10 and 10^n?


As nn gets bigger, which of the two things has the greatest value?
Hence, can you find something that is bigger than your sequence at some point, and is still a null one.

EDIT: I guess this may not be the best advice, given that you need to only use rules.
(edited 12 years ago)
Reply 8
Original post by mootasaurous
If {a} and {b} are null then
{a+b} is null - sum rule
{Ka} is null, K belongs to all real numbers - multiple rule
{ab} is null - product rule
I don't see how to do it using only those rules.

Edit: the "best" I can see requires also: if {a^2} is null than {a} is null.
(edited 12 years ago)
Reply 9
Original post by mootasaurous
I thought about that but according to the multiple rule, the multiplier has to be a real number (i think?) and it wouldnt be multiplying it by a real number, it would be either n^10 or 10^n

or am i being silly


No, I think I was.

10n dominates n10 so you could squeeze it between 2*10n and something maybe?
(edited 12 years ago)
sputum
..
Squeezing isn't one of the "rules" though.

Idea is good though.

Following through: n^10/10^n is null, and 100^n/n! is null, so...
Reply 11
Original post by DFranklin
I don't see how to do it using only those rules.

Edit: the "best" I can see requires also: if {a^2} is null than {a} is null.


is this not the product rule?
Reply 12
Original post by DFranklin
Squeezing isn't one of the "rules" though.

Idea is good though.

Following through: n^10/10^n is null, and 100^n/n! is null, so...


oooh! i think you've got it!

thanks for helping everyone :smile:
"postgraduate"?
Reply 14
Original post by Totally Tom
"postgraduate"?

oh oops
Reply 15
N
Original post by DFranklin
Squeezing isn't one of the "rules" though.

Idea is good though.

Following through: n^10/10^n is null, and 100^n/n! is null, so...


10^n*10^n is not 100^n its 10^2n
Original post by dlynton
N

10^n*10^n is not 100^n its 10^2n
No. 10^n * 10^n = 100^n = 10^2n.
Reply 17
Is {2/n!} a null sequence

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