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Analysis + convergence

  1. Jan 17, 2006 #1
    [tex]f_n(x)= x^n/1 + x^n [/tex]

    does this series converge on the interval [0,1]
    Say if x = 1 then the series is < some epslon , where epslon is > 0
    but if x = 1 then the value for [tex]f_n(x)[/tex] is constant for all n
    but does this still mean the series converges.. even tho the series doesnt get smaller for large values of n.....
    if not then it doesnt converge?
    Last edited: Jan 17, 2006
  2. jcsd
  3. Jan 17, 2006 #2
    read what you posted one more time... your answer is in there somewhere :wink:
  4. Jan 17, 2006 #3


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    Staff Emeritus
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    I assume you mean [itex]f_n(x)= \frac{x^n}{1+ x^n}[/itex] since otherwise (i.e. if you mean [itex]f_n(x)= 2x^n[/itex]) I see no reason for the "1". Also It's not clear whether you mean a "sequence" rather than "series" (which is an infinite sum) since you say "even tho the series doesnt get smaller for large values of n". A sequence doesn't have to "get smaller", just get closer to (or be) a number (an infinites series does not converge it the sequence does not go to 0). fn(0)= 0 for all n and the "constant" sequence 0 certainly converges to 0 (and the infinite sum is also 0!). fn(1)= 1 for all n so the sequence "converges" to 1 but that means that the infinite series [itex]\sum f_n(1)[/itex] does not converge. For x any number between 0 and 1, we can divide both numerator and denominator by xn to get [itex]\frac{1}{1+\frac{1}{x^n}}[/itex]. That sequence also converges to 0 so the series may converge. Since these are all positive values, we can apply the integral test: The series [itex]\sum \frac{x^n}{1+ x^n}[/itex] will converge if and only if the infinite integral [itex]\int_1^\infty \frac{x^y}{1+ x^y} dy[/itex] converges (for 0< x< 1). To integrate that, let u= 1+ xy. Then du= ln(x) xy so
    [tex]\int_1^\infty \frac{x^y}{1+ x^y} dy= \frac{1}{ln x}\int_{1+x}^1\frac{du}{u}= -\frac{ln(1+x)}{ln(x)}[/tex]
    Okay, that exists (and is positive since x< 1), and so the series converges for all x< 1.
    The sequence of functions converges to the discontinuous (at 0) function f(x)= 0 for 0<= x< 1, f(1)= 1.
    The series converges on [0, 1).
    Last edited: Jan 17, 2006
  5. Jan 17, 2006 #4

    D H

    Staff: Mentor

    It's much easier to apply the ratio or comparison test.
    Let [itex] s_n = \sum_{n=0}^\infty f_n(x)[/itex]
    Obviously sn diverges for x>=1 since fn converges to 1 for x=1 and diverges for x>1. Thus we are only concerned with abs(x)<1. Note that each term fn in sn is positive for all x>0 and that the terms alternate in sign for x<0.

    Ratio test: Each term in sn is positive for all x>0.
    sn is a convergent series if [itex]f_{n+1}(x)/f_n(x) < 1[/itex] for all sufficiently large f. The ratio converges to x for x<1, so sn is convergent for [itex]x\in [0,1)[/itex].

    Comparison test: The series [itex]\sum_{n=0}^\infty x^n[/itex] converges for all x in (-1,1). For all x>0, each term [itex]f_n(x) = \frac{x^n}{1+x^n}[/itex] is smaller than [itex]f_n(x) = x^n[/itex]. Thus sn is convergent for [itex]x\in [0,1)[/itex].

    Alternating series test: The sequence [itex]\frac{\lvert x^n\rvert}{1+x^n}[/itex] is monotonically deceasing for [itex]x\in(-1,0)[/itex] and some sufficiently large n. Thus the alternating series \sum_{n=0}^\infty f_n(x), x\in(-1,0)[/itex] is convergent.
    Last edited: Jan 17, 2006
  6. Jan 17, 2006 #5
    It looks easy....

    If [tex]|x| < 1[/tex] it obviously converges to 0.

    If [tex]x = 1[/tex] it is always [tex]1/2[/tex].

    If [tex]x = -1[/tex] it doesn't converge, considering [tex]n[/tex] can take odd values.

    If [tex]|x| > 1, [/tex] [tex]f_n(x)=x^n/(1+x^n) = 1/(1/x^n+1)[/tex], so it converges to 1.
    Last edited: Jan 17, 2006
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