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Thread: Interval of Convergence

  1. #1
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    Interval of Convergence

    I'm trying to determine the interval of convergence for the series:

    $\displaystyle \Sigma_{n=1}^{\infty}\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}{n!}$

    $\displaystyle \lim_{n->\infty}\mid\frac{u_{n+1}}{u_n}\mid=\lim_{n->\infty}\mid\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\c dot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}\mid$

    Using $\displaystyle (n+1)!=n!(n+1)$, this simplifies:

    $\displaystyle =\lim_{n-\infty}\mid\frac{(n+2)x}{(n+1)^2}\mid$

    According to my solution manual, this is incorrect. The limit $\displaystyle \lim_{n->\infty}\mid\frac{u_{n+1}}{u_n}\mid=\lim_{n-\infty}\mid\frac{(n+2)x}{(n+1)}\mid$

    Where am I going wrong?
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  2. #2
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    The calculation should be:

    $\displaystyle \frac{u_{n+1}}{u_n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\c dot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}=\frac{(n+2)x}{(n+1)}$
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    lim |x(n+2)/(n+1)|=|x*lim (n+2)/(n+1)|=|x|*1

    For the convergence you need:

    |x|<1

    Now check for x=1 and x=-1
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  4. #4
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    Quote Originally Posted by elim View Post
    The calculation should be:

    $\displaystyle \frac{u_{n+1}}{u_n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\c dot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}=\frac{(n+2)x}{(n+1)}$
    I know, but I don't see why that's true if $\displaystyle (n+1)!=n!(n+1)$ I obtain:

    $\displaystyle \frac{2\cdot 3 \cdot 4\cdot\cdot\cdot (n+2)x^{n+1}}{n!(n+1)}\frac{n!}{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)x^n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}\frac{(n+2)x}{(n+1)(n+1)}=\frac{(n+2)x}{(n+1)^2 }$

    So I cancel out the $\displaystyle 2\cdot 3\cdot 4\cdot\cdot\cdot x^nn!$

    Notice that there are TWO $\displaystyle (n+1)$ factors in the denominator. Where does that second factor go?
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  5. #5
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    Quote Originally Posted by Also sprach Zarathustra View Post
    lim |x(n+2)/(n+1)|=|x*lim (n+2)/(n+1)|=|x|*1

    For the convergence you need:

    |x|<1

    Now check for x=1 and x=-1
    I understand from that point, my problem is the simplification of $\displaystyle \frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\c dot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}$.

    See my last reply to elim
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  6. #6
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    Quote Originally Posted by adkinsjr View Post
    I know, but I don't see why that's true if $\displaystyle (n+1)!=n!(n+1)$ I obtain:

    $\displaystyle \frac{2\cdot 3 \cdot 4\cdot\cdot\cdot (n+2)x^{n+1}}{n!(n+1)}\frac{n!}{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)x^n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}\frac{(n+2)x}{(n+1)(n+1)}=\frac{(n+2)x}{(n+1)^2 }$

    So I cancel out the $\displaystyle 2\cdot 3\cdot 4\cdot\cdot\cdot x^nn!$

    Notice that there are TWO $\displaystyle (n+1)$ factors in the denominator. Where does that second factor go?
    The first equality is incorrect. How did you get $\displaystyle 2 \cdot 3 \cdot 4 \cdots n! \cdot x^n$ in the numerator? It should be $\displaystyle 2 \cdot 3 \cdot 4 \cdots n \cdot x^n$ - you have an extra factorial there.
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  7. #7
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    I think I've got it now.

    The sequence can be written as $\displaystyle \frac{(n+1)!x^n}{n!}$
    $\displaystyle \frac{u_{n+1}}{u_n}=\frac{(n+1)!(n+2)x^{n+1}n!}{(n +1)!x^n(n+1)!}=\frac{(n+2)x}{n+1}$

    That seems right.
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  8. #8
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    Quote Originally Posted by Defunkt View Post
    The first equality is incorrect. How did you get $\displaystyle 2 \cdot 3 \cdot 4 \cdots n! \cdot x^n$ in the numerator? It should be $\displaystyle 2 \cdot 3 \cdot 4 \cdots n \cdot x^n$ - you have an extra factorial there.
    I don't see what you mean. The numerators are both the same (as far as I can tell), just rearranged.

    $\displaystyle 2\cdot 3\cdot 4\cdot (n+2)x^{n+1}n!=2\cdot 3\cdot 4\cdot n!x^n(n+2)x$
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  9. #9
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    Oh, you're right; my bad. Looks like you just missed the $\displaystyle n+1$ term from the numerator of the first fraction.
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  10. #10
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    The first thing I would do is simplify $\displaystyle \frac{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)}{n!}$ to $\displaystyle \frac{(n+1)!}{n!}= n+1$.

    The series you have is simply $\displaystyle \sum_{n=1}^\infty (n+1)x^n$.
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    hmmm.... I don't know about this simplification...
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  12. #12
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    Quote Originally Posted by Also sprach Zarathustra View Post
    hmmm.... I don't know about this simplification...
    What do you not understand about that?
    It is completely obvious.
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  13. #13
    MHF Contributor Also sprach Zarathustra's Avatar
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    Maybe it because of the night, I don't know! Anyway...

    {2*4*6...*(n+1)}*{1*3*5*7*...*n} =(n+1)!


    Edit: ooops(again!) I have not noticed that is not only the even numbers!

    Sorry!
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  14. #14
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    Quote Originally Posted by HallsofIvy View Post
    The first thing I would do is simplify $\displaystyle \frac{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)}{n!}$ to $\displaystyle \frac{(n+1)!}{n!}= n+1$.

    The series you have is simply $\displaystyle \sum_{n=1}^\infty (n+1)x^n$.
    Yes, I realize I should have done that. I was foolishly jumping right into the ratio test, without looking for any simplifications in the series itself. I wasted too much time on this, lol.
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  15. #15
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Maybe it because of the night, I don't know! Anyway...

    {2*4*6...*(n+1)}*{1*3*5*7*...*n} =(n+1)!


    Edit: ooops(again!) I have not noticed that is not only the even numbers!

    Sorry!
    A product of even numbers can be reduced as:
    $\displaystyle 2(4)(6)\cdot\cdot\cdot(2n-2)(2n)= [(1)(2)][(2)(2)][(3)(2)]\cdot\cdot\cdot[(n-1)(2)][(n)(2)]$$\displaystyle = n!2^n$.

    A product of odd numbers is more complicated- you have to put the even numbers in:
    $\displaystyle (1)(3)(5)\cdot\cdot\cdot (2n-1)(2n+1)= \frac{(1)(2)(3)(4)(5)\cdot\cdot\cdot(2n-1)(2n)(2n+1)}{2(4)(6)\cdot\cdot\cdot(2n-2)(2n)}= \frac{(2n+1)!}{2^n n!}$.
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