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Thread: limit of a series

  1. #1
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    limit of a series

    $\displaystyle y_n=\frac{t^{n+1}}{(n+1)!}+2\sum_{k=0}^{n}\frac{t^ {k}}{k!}-t-1$
    what is the limit of this series when n goes to infinity
    ?
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    $\displaystyle y_n=\frac{t^{n+1}}{(n+1)!}+2\sum_{k=0}^{n}\frac{t^ {k}}{k!}-t-1$
    what is the limit of this series when n goes to infinity
    ?
    For any fixed $\displaystyle t \in \mathbb{R}$

    $\displaystyle \displaystyle \lim_{n \to \infty}\frac{t^{n+1}}{(n+1)!}=0$


    $\displaystyle \displaystyle \lim_{n \to \infty}\sum_{k=0}^{n}\frac{t^k}{k!}=e^t$

    This is the Taylor series of $\displaystyle e^t$ centered a 0.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    For any fixed $\displaystyle t \in \mathbb{R}$

    $\displaystyle \displaystyle \lim_{n \to \infty}\frac{t^{n+1}}{(n+1)!}=0$

    why its zero i dont know whats t?
    the denominetor could be zero too
    $\displaystyle \displaystyle \lim_{n \to \infty}\sum_{k=0}^{n}\frac{t^k}{k!}=e^t$

    This is the Taylor series of $\displaystyle e^t$ centered a 0.
    how to prove that its exponent series
    i cant write "its the same"

    what about -t -1
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  4. #4
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    $\displaystyle \displaystyle f(t)=e^t \implies \frac{d^n}{dt^n}f(t)=e^t$ for all $\displaystyle n$.
    $\displaystyle \displaystyle \frac{d^n}{dt^n}f(t)\bigg|_{t=0}=1$ for all $\displaystyle n$

    Using the definition of the Taylor series gives

    $\displaystyle \displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}t^n=\sum_{ n=0}^{\infty}\frac{t^n}{n!}$

    Since $\displaystyle -t-1$ does not depend on n what is the limit

    $\displaystyle \displaystyle \lim_{n \to \infty}(-t-1)=(-t-1)\lim_{n \to \infty}1=?$
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