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Can't find formula of a serie.

  1. May 8, 2009 #1
    A problem from my old Calculus book I can't solve...

    1. The problem statement, all variables and given/known data

    Find the formula of this serie knowing its first five terms:

    [tex]1 +\left(\frac{2}{5}\right)^{2} +\left(\frac{3}{8}\right)^{3} +\left(\frac{4}{11}\right)^{4} +\left(\frac{5}{14}\right)^{5} + \cdots[/tex]

    2. Relevant formulas

    If the first term was 1/2 then the formula would simply be

    [tex]\sum_{i = 1}^{\infty}\left(\frac{i}{2+3(i-1)}\right)^{i}[/tex]

    but the first term being 1, I can't see how to do it...

    (Maybe the first term IS 1/2 and it got missprinted in the book ?)

    Also, the answer is not given in the book.

    3. The attempt at a solution

    I've tried these as possible values for a first item equal to 1 without any success:




    So, is there a formula for this serie with first term = 1 ?

    Thanks !
  2. jcsd
  3. May 9, 2009 #2
    could you do 1 + [tex]
    \sum_{i = 2}^{\infty}\left(\frac{i}{3i-1)}\right)^{i}
    [/tex] ? I can't think of a series that fits there without just not counting the "first" term.
  4. May 9, 2009 #3


    Staff: Mentor

    How about this?
    [tex]1/2 +\left(\frac{1}{2}\right)^{1} + \left(\frac{2}{5}\right)^{2} +\left(\frac{3}{8}\right)^{3} +\left(\frac{4}{11}\right)^{4} +\left(\frac{5}{14}\right)^{5} + \cdots[/tex]

    Now you can write this as 1/2 + the summation you had, starting at i = 1, though. I don't see any problem with doing that.
  5. May 9, 2009 #4
    Thanks Hogger and Mark44.

    I think you got it !

    I needed the formula of the serie to figure out if it converge or not.
    Substracting 1/2 from it made me remember about this theorem from the book:

    "Convergence or divergence of a serie is not affected by adding or substracting a finite number of terms from it."

    So, and if I get all this right, convergence or divergence of these series will be the same:

    The serie from the book:

    [tex]1 +\left(\frac{2}{5}\right)^{2} +\left(\frac{3}{8}\right)^{3} +\left(\frac{4}{11}\right)^{4} +\left(\frac{5}{14}\right)^{5} + \cdots[/tex]

    The serie minus 1/2:

    [tex]\left(\frac{1}{2}\right)^{1} +\left(\frac{2}{5}\right)^{2} +\left(\frac{3}{8}\right)^{3} +\left(\frac{4}{11}\right)^{4} +\left(\frac{5}{14}\right)^{5} + \cdots[/tex]

    The second serie brings us back to the Relevant formula of my original post and I know how to find-out if it converge or not.

    If it does, the serie from the book does to.

    Is this right ?

    Thanks !
  6. May 10, 2009 #5


    Staff: Mentor

    Yes, this is right.

    BTW, there is no such word in English as "serie." The word is series, which is both singular and plural.
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