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Convergence of a general series

  1. Jul 5, 2011 #1
    This problem has been bothering me for some time. Any thoughts or insights are greatly appreciated.

    Consider a function, f, with continuous third derivative on [-1,1]. Prove that the series
    [itex]\sum^{\infty}_{n=1} (nf(\frac{1}{n})-nf(-\frac{1}{n}) - 2\frac{df}{dn}(0))[/itex] converges.

    Thanks in advance for any help!
  2. jcsd
  3. Jul 6, 2011 #2
    I think the statement is false. Try with f(x) = x.
  4. Jul 6, 2011 #3


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    You also cannot have "[itex]df/dn[/itex]" since f is not a function of n.

    Perhaps you meant
    [tex]\sum_{n=1}^\infty\left(nf(1/n)+ nf(-1/n)- 2\frac{df}{dx}(0)\right)[/tex]
  5. Jul 6, 2011 #4

    I like Serena

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    Uhh, the statement holds true for f(x)=x (assuming df/dn is actually df/dx as HoI suggested). :uhh:

    @HoI: You flipped a sign (typo I presume?). For your current expression f(x)=x will not hold.
  6. Jul 6, 2011 #5
    As for a proof, you might want to take the Taylor expansion of f at 0. A lot of terms will drop and you'll see fun things happening.
  7. Jul 6, 2011 #6
    I found the problem in the form I presented, though the derivative was with respect to x (as HoI pointed out). If I expand f as a taylor series, I think I see the fun of which you speak.

    EDIT: All we know about f is that its third derivative is continuous. Does this effect the taylor series (Is it invalid to use derivatives at 0 of order > 3 if we are unsure if they even exist?)?
    Last edited: Jul 6, 2011
  8. Jul 6, 2011 #7

    I like Serena

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    You can't use the full Taylor series expansion (you don't even know it converges).

    However, you can use Taylor up to 4 terms and use the term with the second or third derivative to define a remainder term that bounds the function result.
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