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Math Help - Power series and convergence

  1. #1
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    Power series and convergence

    Could somebody explain me how to find the convergence and the radius of the convergence for the following power series?
    Attached Thumbnails Attached Thumbnails Power series and convergence-kkk.jpg  
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  2. #2
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    Did you try the ratio test. When you do and take the limit,

    \frac{a_{n+1}}{a_{n}}=\frac{ln(k+2)(x+1)^{k+1}}{k+  2}\cdot\frac{k+1}{ln(k+1)(x+1)^{k}}

    \lim_{k\rightarrow{\infty}}\frac{a_{n+1}}{a_{n}}=|  x+1|

    I may as well finish.

    -1<x+1<1

    -2<x<0

    Diverges if x<-2 or of x>0

    Check the endpoints:

    If we enter in x=-2 into our series we get \sum_{k=1}^{\infty}\frac{(-1)^{k}ln(k+1)}{k+1}, converges. (converges to \approx{-.15986...})

    If we enter in x=0, we get \sum_{k=1}^{\infty}\frac{ln(k+1)}{k+1}, diverges.

    So, the interval of convergence is [-2,0) with radius of 1

    My assessment agrees with MS except are you sure when x=0 it converges?.
    Last edited by galactus; June 1st 2008 at 01:48 PM.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by totalnewbie View Post
    Could somebody explain me how to find the convergence and the radius of the convergence for the following power series?
    Galactus's method is totally right, but just so you get more than one method

    Root test

    \sum_{n=1}^{\infty}\frac{\ln(n+1)(x+1)^n}{n+1}

    So we need to find all values of x such that

    \lim_{n\to\infty}\bigg|\frac{\ln(n+1)(x+1)^n}{n+1}  \bigg|^{\frac{1}{n}}<1

    So we see that

    \lim_{n\to\infty}\bigg|\frac{\ln^{\frac{1}{n}}(n+1  )(x+1)}{(n+1)^{\frac{1}{n}}}\bigg|<1\Rightarrow|x+  1|<1

    Or in other words

    x+1<1\Rightarrow{x<0}

    or

    x+1>-1\Rightarrow{x>-2}

    Now we need to check endpoint convergence

    at x=0 we have

    divergence by integral test

    at x=-2 we have

    \sum_{n=0}^{\infty}\frac{ln(n+1)(-1)^n}{n+1}

    and since \exists{N}\backepsilon\forall{n>N},a_{n+1}<a_n

    and also since \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{\ln(n+  1)}{n+1}=0

    this series converges by AST

    Therefoe IOC is [-2,0)


    EDIT: For purely supplemetary purposes

    Let L=\lim_{n\to\infty}\ln^{\frac{1}{n}}(n+1)\Rightarr  ow{\ln(L)=\lim_{n\to\infty}\frac{\ln(\ln(n+1))}{n}  }

    For the right hand limit you can use many methods, L'hopital's is most preferable here

    \lim_{n\to\infty}\frac{\ln(\ln(n+1))}{n}=\lim_{n\t  o\infty}\frac{\frac{1}{(n+1)\ln(n+1)}}{1}=0

    So we have that

    \ln(L)=\lim_{n\to\infty}\frac{\ln(\ln(n+1))}{n}=0\  Rightarrow{L=e^0=1}

    Use the same method to compute L=\lim_{n\to\infty}(n+1)^{\frac{1}{n}}
    Last edited by Mathstud28; June 1st 2008 at 01:43 PM.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Did you try the ratio test. When you do and take the limit,

    \frac{a_{n+1}}{a_{n}}=\frac{ln(k+2)(x+1)^{k+1}}{k+  2}\cdot\frac{k+1}{ln(k+1)(x+1)^{k}}

    \lim_{k\rightarrow{\infty}}\frac{a_{n+1}}{a_{n}}=|  x+1|

    I may as well finish.

    -1<x+1<1

    -2<x<0

    Diverges if x<-2 or of x>0

    Check the endpoints:

    If we enter in x=-2 into our series we get \sum_{k=1}^{\infty}\frac{(-1)^{k}ln(k+1)}{k+1}, converges. (converges to \approx{-.15986...})

    If we enter in x=0, we get \frac{ln(k+1)}{k+1}, diverges.

    So, the interval of convergence is [-2,0) with radius of 1

    My assessment agrees with MS except are you sure when x=0 it converges?.
    Thanks so much! I fixed it before you posted that , the entire time I kept typing x instead of x+1, therefore when I saw x=0 I said a sequence of all zero terms, I am sorry for my stupid error , I knew it would bite me sooner or later

    Just to show once again, now that I see it is (x+1)^n and not x^n

    x=0 gives

    \sum_{n=1}^{\infty}\frac{ln(n+1)}{n+1}

    and as we stated earlier the existence of eventual monotonic decrease and all the terms are positve we can apply the integral test


    \int_1^{\infty}\frac{ln(n+1)}{n+1}=\frac{1}{2}\ln^  2(n+1)\bigg|_1^{\infty}=\infty

    therefore divergent
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