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Math Help - Taylor remainder

  1. #1
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    Taylor remainder

    I was finding the Taylor expansion for f(x)=\ln x at x=1 and I've found that f^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{x^n} but I need to show that the remainder tends to zero as n\to\infty, my problem is that I don't know what's the form of the remainder and how to show that it tends to zero.
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  2. #2
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    Quote Originally Posted by Astrid View Post
    I was finding the Taylor expansion for f(x)=\ln x at x=1 and I've found that f^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{x^n} but I need to show that the remainder tends to zero as n\to\infty, my problem is that I don't know what's the form of the remainder and how to show that it tends to zero.

    I think that as you tend to infinity (-1)^{n+1} makes no difference it just changes the sign so igrore this, then you have \frac{(n-1)!}{x^n} as you know that x^{n} tends to infinite quicker than (n-1)! tends to infinit you get \frac{small infinite}{big infinite}

    hope this helps
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  3. #3
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    no, it doesn't, sorry.

    what's the formula of the remainder? how to compute it's limit?
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  4. #4
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    Quote Originally Posted by Astrid View Post
    no, it doesn't, sorry.

    what's the formula of the remainder? how to compute it's limit?

    Okay if u think about any fraction if u have a large number on the bottem and a small number on the top you end up with ur problem tending to 0.
    N sorry dont know what the formula for the remainder is.

    bobisback
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