- #1

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## Homework Statement

x^n / n! -> 0 for any value of x and n -> 0

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- Thread starter burhan1
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- #1

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x^n / n! -> 0 for any value of x and n -> 0

- #2

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What have you tried to do so far?

- #3

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hey wait maybe i could prove that |x^n|/(n!) -> 0 and use the absolute value rule...

- #4

Mark44

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I don't think you have stated the problem correctly. Isn't this limit as n --> ∞?## Homework Statement

x^n / n! -> 0 for any value of x and n -> 0

- #5

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I don't think you have stated the problem correctly. Isn't this limit as n --> ∞?

I think so too. The limit of x

As for x

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- #6

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Oops yeah n -> infinity. sorry about that =/

- #7

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- #8

HallsofIvy

Science Advisor

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- #9

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im a first year undergraduate i have nothing on me panicking like crazy xD

and that didnt help TT_TT

and that didnt help TT_TT

- #10

lurflurf

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Prove that for n large

(2/n)^{n}<1/n!<(3/n)^{n}

(2/n)

- #11

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Show that [x^(n+1)/(n+1)!] < [x^n/n!] when n>>x.

Then [x^(n+1)/(n+1)!] / [x^n/n!] = x/(n+1) = 0 when n→∞.

Then [x^(n+1)/(n+1)!] / [x^n/n!] = x/(n+1) = 0 when n→∞.

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- #12

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heres what im trying:

Prove that a_n = (x^n/n!) -> 0 for all x

Choose ε > 0. Then there exists for all N ,which belongs to natural numbers, n>N and |a_n| < ε

so, |(x^n/n!)| < ε => |x^n| < n!*ε (since n! is positive the absolute sign is not significant?)

so |x|/[ε^(1/n)] < [n!^(1/n)].

so choose any natural number N s.t. N > |x|/[ε^(1/n)]

If n>N then x^n/n! < x^N/N! < ε

which means a_n -> 0.

is this correct in any sense? =S

- #13

lurflurf

Homework Helper

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lim (a x/n)

because

(2/n)

implies

x

- #14

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no that is the next question lurf :P

i've done it guys. thanks anyway for the help!

i've done it guys. thanks anyway for the help!

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