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Ratio Test Question

I think this is right I'm just not 100% sure if I'm using the ratio test correctly.

When does the series n!x^n converge?

By ratio test we get (n+1)!x^(n+1)/(n!x^n)=nx which diverges unless x=0 so the series only converges when x=0

Is this right?
Reply 1
Avatar for Smaug123
Smaug123
15
Original post by james22
I think this is right I'm just not 100% sure if I'm using the ratio test correctly.

When does the series n!x^n converge?

By ratio test we get (n+1)!x^(n+1)/(n!x^n)=nx which diverges unless x=0 so the series only converges when x=0

Is this right?

Strictly, that argument doesn't actually show that the series converges at x=0; it shows that if it converges at x, then x = 0. Otherwise, entirely correct :smile: you just need an extra line saying "since n! x^n = 0 for x=0, the sum is 0 for x=0 and hence converges for x=0". (since the ratio test can't be used if the denominator's 0.)
Original post by james22
I think this is right I'm just not 100% sure if I'm using the ratio test correctly.

When does the series n!x^n converge?

By ratio test we get (n+1)!x^(n+1)/(n!x^n)=nx which diverges unless x=0 so the series only converges when x=0

Is this right?


I get that it converges when x is less than or equal to 1/(n+1)

since you'll get that (n+1)!x^(n+1)/(n!x^n)= (n+1)x

to converge (n+1)x < 1 by the ratio test

and so x < 1/(n+1)

and lim as n--> infinity is just 1/(infinity + 1) which ----> 0

and so x = 0
(edited 11 years ago)
Reply 3
Avatar for Noble.
Noble.
17
Do you mean the series (n!xn)\left(n!x^n \right) or ∑n=0∞n!xn\displaystyle\sum_{n=0}^{\infty} n!x^n?

The series only converges if ∣x∣<1|x|<1

Clearly then the sum can only possibly converge if ∣x∣<1|x|<1 (because for the sum to potentially converge we must have n!xn→0n!x^n \rightarrow 0 as n→∞n \rightarrow \infty)

However, the ratio test shows that

lim⁡n→∞∣(n+1)!xn+1n!xn∣=lim⁡n→∞∣(n+1)x∣\displaystyle\lim_{n \rightarrow \infty} \left|\dfrac{(n+1)!x^{n+1}}{n!x^n}\right| = \displaystyle\lim_{n \rightarrow \infty} \left| (n+1)x \right|

So, for convergence we must have that lim⁡n→∞∣(n+1)x∣<1\displaystyle\lim_{n \rightarrow \infty} |(n+1)x| < 1 which is not possible unless x=0x = 0 so, as you say, the sum doesn't converge unless x=0x=0
(edited 11 years ago)
Reply 4
Avatar for Noble.
Noble.
17
Original post by Black_Materia_
I get that it converges when x is less than or equal to 1/(n+1)

since you'll get that (n+1)!x^(n+1)/(n!x^n)= (n+1)x

to converge (n+1)x < 1 by the ratio test

and so x < 1/(n+1)


You're forgetting that n→∞n \rightarrow \infty which implies x=0x=0 as the only case when the sum converges.
Reply 5
Avatar for james22
james22
OP
16
Thanks, I meant the sum not the sequence..

BTW x cannot depend on n is this, x is just a number.
Reply 6
Avatar for Noble.
Noble.
17
Original post by james22
Thanks, I meant the sum not the sequence..

BTW x cannot depend on n is this, x is just a number.


Ok, thought you meant the sum.

Well, for it to converge you must have:

lim⁡n→∞∣(n+1)x∣<1\displaystyle\lim_{n \rightarrow \infty} \left| (n+1)x \right| < 1 by the ratio test, but this is only possible if x=0x=0 so it does not converge for any non-zero xx

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