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    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?
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    (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.)
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    (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
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    Do you mean the series \left(n!x^n \right) or \displaystyle\sum_{n=0}^{\infty} n!x^n?

    The series only converges if |x|&lt;1

    Clearly then the sum can only possibly converge if |x|&lt;1 (because for the sum to potentially converge we must have n!x^n \rightarrow 0 as n \rightarrow \infty)

    However, the ratio test shows that

    \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 \displaystyle\lim_{n \rightarrow \infty} |(n+1)x| &lt; 1 which is not possible unless x = 0 so, as you say, the sum doesn't converge unless x=0
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    (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 \rightarrow \infty which implies x=0 as the only case when the sum converges.
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    Thanks, I meant the sum not the sequence..

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

    \displaystyle\lim_{n \rightarrow \infty} \left| (n+1)x \right| &lt; 1 by the ratio test, but this is only possible if x=0 so it does not converge for any non-zero x
 
 
 
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