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Methods for convergence divergence

  1. Nov 4, 2004 #1
    I have [tex]\sum_{n=1}^\infty{\frac{1}{n^2+n+1}}[/tex] and I need to show that it converges or diverges. I choose to do the comparison test making [tex]A_n=\sum_{n=1}^\infty{\frac{1}{n^2+n+1}}[/tex] and[tex] B_n=\sum_{n=1}^{\infty}\frac{1}{n^2+n}[/tex] so far so good? Okay well [tex] \lim_{n\rightarrow0}B_n=0[/tex] so does [tex]A_n[/tex] converge...i see that the upper limit of [tex]A_n[/tex] would turn out to be [tex]0[/tex] what does this mean...is it valid to use the rule I used?

    what if did [tex]\int_{1}^{\infty}\frac{1}{x^2+x+1}dx[/tex] is that possible or is there no need?
     
    Last edited: Nov 4, 2004
  2. jcsd
  3. Nov 4, 2004 #2

    arildno

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    Here, you are confusing indexes!!!

    Your expressions have no meaning; what convergence in this case means, is that the SEQUENCE OF PARTIAL (FINITE!!!) SUMS CONVERGE.

    Hence, your index in that sequence SHOULD BE THE UPPER LIMIT VALUE IN THE SUM, not the summation index you've used!!

    We therefore have the partial sum:
    [tex]A_{N}=\sum_{n=1}^{N}\frac{1}{n^{2}+n+1}[/tex]

    Our question is therefore:
    Does the sequence [tex]A_{N}[/tex] converge as [tex]N\to\infty[/tex] ??

    To help you on your way, note that:
    [tex]A_{N}\leq{B}_{N},B_{N}=\sum_{n=1}^{N}\frac{1}{n^{2}}[/tex]
     
  4. Nov 4, 2004 #3
    thats a p series and its greater than one...easy...so [tex] A_n[/tex] must converge...is that my answer?
     
  5. Nov 5, 2004 #4

    arildno

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    That's right.
     
  6. Nov 5, 2004 #5

    HallsofIvy

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    You are also confusing "series" and "sequences". If
    [tex] B_n=\sum_{n=1}^{N}\frac{1}{n^2+n}[/tex]

    then [tex] \lim_{n\rightarrow0}B_n=0[/tex]
    is NOT true. What is true is that the SEQUENCE {Bn[/b]} goes to 0 and that tells you nothing about the series! (If the sequence did NOT converge to 0, that would tell you that the series does not converge.)
     
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