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Ratio Test Question watch

1. 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?
2. (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 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.)
3. (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
4. Do you mean the series or ?

The series only converges if

Clearly then the sum can only possibly converge if (because for the sum to potentially converge we must have as )

However, the ratio test shows that

So, for convergence we must have that which is not possible unless so, as you say, the sum doesn't converge unless
5. (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 which implies as the only case when the sum converges.
6. Thanks, I meant the sum not the sequence..

BTW x cannot depend on n is this, x is just a number.
7. (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:

by the ratio test, but this is only possible if so it does not converge for any non-zero

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