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Series and sequence

  1. Apr 23, 2005 #1
    [tex]An = \frac{5n}{12n+5}[/tex]

    For both of the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise.


    a.) The series n=1 to infinity(i dont know how to make the 'sum of' sign)
    (An) i'm having trouble with this one, there seems to be no 'common ratio', so does this mean it's divergent? [tex]A_1 = 5/17, A_2 = 10/29, A_3 = 15/41 [/tex]

    b.) the squence [tex]A_n[/tex]. well for this one as n-> infinity, the limit should be 5/12
     
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  3. Apr 23, 2005 #2

    OlderDan

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    Your b) looks good and should tell you the answer to a) What is 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 +. . .
     
  4. Apr 23, 2005 #3

    James R

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    If

    [tex]A_n = \frac{5n}{12n+5}[/tex]

    then

    [tex]\lim_{n \rightarrow \infty} A_n = \lim_{n \rightarrow \infty} \frac{5}{12+5/n}[/tex]
    [tex]=\frac{5}{12+(\lim_{n \rightarrow \infty} 5/n)} = 5/12[/tex]
     
  5. Apr 23, 2005 #4

    HallsofIvy

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    Hint for a) In order for a series such as [itex]\Sigma_1^{\infinity}A_n[/itex] to converge, it is necessary that the sequence {An} converge to 0.
     
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