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Alternating series something dissapeared.

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi,
    The question wanted to know if the alternating series converges or diverges.

    $$A_n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}$$

    2. Relevant equations



    3. The attempt at a solution[/b
    You can see it here .

    http://www.calcchat.com/book/Calculus-ETF-5e/ chapter 9, section 5, question 11
    So I understand the process I just want to know why they ignore the $$ (-1)^{n+1}$$ in the numerator all of a sudden it just becomes 1.


    I understand that it will be either 1 or -1 depending on the value of n. But why can they just ignore if it was -1 and say it is 1 and take the limit? I don't get it.
     
  2. jcsd
  3. Jul 7, 2013 #2

    Office_Shredder

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    The statement of theorem 9.14 is going to start with: suppose you have a sum of the form
    [tex] \sum (-1)^{n+1} a_n [/tex]

    Where the an are positive and decreasing.

    So the theorem says if you have an alternating series, you can take the absolute value of the terms to show that it converges (usually)
     
  4. Jul 7, 2013 #3
    Theorem 9.14

    Let $$A_n >0 $$.
    The alternating series

    $$\sum (-1)^{n} a_n$$ and $$\sum (-1)^{n+1} a_n$$
    converge if the following two conditions are met.

    1. lim --> ∞ $$A_n = 0$$ 2. $$A_n+1 <= A_n for all n$$
     
  5. Jul 7, 2013 #4
    Ahh crap for 2 I meant A_(n+1) <= A_n for all n

    I suck at the writing the equations. But yeah no abs anywhere in that theorm
     
  6. Jul 8, 2013 #5
    All the theorem is trying to say is that if the [itex]a_n[/itex] sequence is decreasing, then the alternating series is convergent.

    In this case, the series is convergent, and more specifically, it converges to [itex]1-\log(2)[/itex].
     
  7. Jul 8, 2013 #6

    Curious3141

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    Theorem 9.14 is a statement of a test for convergence called the "alternating series test", also known as the Leibniz criterion. It applies ONLY to alternating series, which have terms that oscillate between positive and negative from one term to the next. It does NOT apply to series where all the terms are positive or all the terms are negative, or even when the positive and negative terms don't "neatly" alternate from one term to the next.

    This criterion requires that (for such an alternating series to converge) when you disregard the sign and just take the absolute value (only the numerical magnitude) of each term, these values go monotonically to zero at the limit.

    Read that carefully. You need to prove that there's a steady trend where the next value is lower numerically than the one before, and that the "last term" (limit) goes to zero. Theorem 9.14 puts that in mathematical notation.

    When you list the absolute values of the terms in this series, they in fact form part of the harmonic sequence omitting 1, viz.: ##\displaystyle \frac{1}{2}, \frac{1}{3}, \frac{1}{4},...,\frac{1}{k},...##. The proof is just showing that it's always true that each term in this sequence is smaller than the one before; and further that the limit of the sequence goes to zero. Both of these are trivial, and you should easily be able to see these.

    This should be sufficient to prove that the alternating series formed from these terms converges (based on the Leibniz criterion). However, even though we brought in the absolute values of the terms, don't get confused into calling the series "absolutely convergent", because it is NOT absolutely convergent. If you take all those absolute values and add them together, you get the harmonic series minus one, i.e. ##\displaystyle H - 1##, which is clearly divergent. It is purely because this is an alternating series formed with those same terms that you can use a "weaker" criterion (like the Leibniz criterion) to show convergence.
     
    Last edited: Jul 8, 2013
  8. Jul 8, 2013 #7

    vanhees71

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    It's convergent, because the squence [itex]a_n=1/(n+1)[/itex] is monotonically decreasing and going to 0 for [itex]n \rightarrow \infty[/itex]. Then according to the Leibniz criterion the alternating series converges.

    To get the value, use the trick of generating functions and define
    [tex]f(x)=\sum_{n=1}^{\infty} (-x)^n.[/tex]
    This series is convergent for [itex]|x|<1[/itex]. Since it's a geometric series you can sum it explicitly
    [tex]f(x)=\frac{1}{x+1}-1.[/tex]
    Now you get your series by integrating [itex]f(x)[/itex] from 0 to 1 (argue why!):
    [tex]\int_0^1 \mathrm{d} x f(x)=[\ln(x+1)-x]_{x=0}^{x=1}=\ln 2-1.[/tex]
     
  9. Jul 9, 2013 #8
    Correct logic, but messes up towards the end and gives the negative of the actual answer. It shouldn't take a genius to see that your answer is negative while the series is supposed to turn out positive.

    A way of going with it is to start with the function
    [tex]f(x) = \sum_{n=1}^{\infty} x^n = \frac{1}{1-x}-1[/tex]
    Integrating this function gives a new function
    [tex]F(x) = \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} = -\log(1-x)-x+C[/tex]
    You can find the constant of integration by setting [itex]x=0[/itex], which gives [itex]C=0[/itex]. Hence, you have:
    [tex]\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} = -x-\log(1-x)[/tex]
    Now set [itex]x=-1[/itex] on both sides to get the answer [itex]1-\log(2)[/itex].

    Note that this only finds the value of the series if it actually converges. There are cases in which methods like this are applied to divergent series to find a "value" for them. This is called regularization. To show that the series actually has this value, you would first need to prove that it is convergent.
     
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