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Math Help - Dealing with Alternating Series

  1. #1
    Member RedBarchetta's Avatar
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    Dealing with Alternating Series

    Is there any particular method to use when dealing with these? Let me explain my problem. For example:

    Here are the rules we know:

    1. The Alternating Series Test

    If the series

    <br />
\sum\limits_{n = 1}^\infty  {( - 1)^{n + 1} u_n  = u_1  - u_2  + u_3  - u_4  +  \cdot  \cdot  \cdot } <br />

    converges if all three of the following conditions are satisfied:

    a. <br />
u_n <br />
is positive for all n.

    b. <br />
u_n  \geqslant u_{n + 1} <br />
for all <br />
n \geqslant N<br />
, for some integer N.

    c. <br />
u_n  \to 0<br />

    2. Absolute Convergence

    Given <br />
\sum\limits_{n = 1}^\infty  {a_n } <br />
if <br />
\sum\limits_{n = 1}^\infty  {|a_n } |<br />
converges, we say that <br />
\sum\limits_{n = 1}^\infty  {a_n } <br />
converges absolutely.

    3. Conditional Convergence

    If <br />
\sum\limits_{n = 1}^\infty  {|a_n } |<br />
diverges, but <br />
\sum\limits_{n = 1}^\infty  {a_n } <br />
converges, then we say that <br />
\sum\limits_{n = 1}^\infty  {a_n } <br />
converges conditionally.

    Example One:

    <br />
\sum\limits_{n = 1}^\infty  {( - 1)^{n + 1} \frac{{3 + n}}<br />
{{5 + n}}} <br />

    So where do I start with an alternating series? In this instance, we know <br />
{\frac{{3 + n}}<br />
{{5 + n}}}<br />
is an increasing function, so it fails part b of the alternating series test. Also <br />
{\mathop {\lim }\limits_{n \to \infty } \frac{{3 + n}}<br />
{{5 + n}}}<br />
approaches one as n goes to infinity. So this alternating series fails the alternating series test completely. So therefore it diverges?

    Example Two
    <br />
\sum\limits_{n = 1}^\infty  {( - 1)^{n + 1} } \frac{n}<br />
{{n^3  + 1}}<br />

    What about this one? I think this one meets of the conditions of the alternating series test. So then, if a series meets all the conditions, do I then use the conditional and absolute convergence tests? or do I just leave it at "converges by alternating series test" What is a good rule of thumb for these series?
    Last edited by RedBarchetta; October 26th 2008 at 01:42 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    Example One:

    <br />
\sum\limits_{n = 1}^\infty  {( - 1)^{n + 1} \frac{{3 + n}}<br />
{{5 + n}}} <br />

    So where do I start with an alternating series? In this instance, we know <br />
{\frac{{3 + n}}<br />
{{5 + n}}}<br />
is an increasing function, so it fails part b of the alternating series test. Also <br />
{\mathop {\lim }\limits_{n \to \infty } \frac{{3 + n}}<br />
{{5 + n}}}<br />
approaches one as n goes to infinity. So this alternating series fails the alternating series test completely. So therefore it diverges?

    I mean, what is a good rule of thumb for these series? If an alternating series fails one of the conditions of the alternating series test, the series automatically diverges?

    Thank you!
    the alternating series test cannot be applied here, since the function is increasing.

    it diverges by the test for divergence, since the limit you mentioned does not go to 0
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  3. #3
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Jhevon View Post
    the alternating series test cannot be applied here, since the function is increasing.

    it diverges by the test for divergence, since the limit you mentioned does not go to 0
    So if one of the conditions aren't met by the alternating series test, then you can't use it at all? It doesn't mean that if one condition isn't met then the alternating series diverges?

    So:

    If <br />
\mathop {\lim }\limits_{n \to \infty } u_n  \ne 0<br />
(i.e. <br />
u_n <br />
diverges) then <br />
\sum\limits_{n = 1}^\infty  {( - 1)^{n + 1} } u_n <br />
diverges?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    So if one of the conditions aren't met by the alternating series test, then you can't use it at all? It doesn't mean that if one condition isn't met then the alternating series diverges?
    yes

    but your conditions aren't right. for example, it being a decreasing function is a condition to use the test, but the limit going to zero isn't


    So:

    If <br />
\mathop {\lim }\limits_{n \to \infty } u_n  \ne 0<br />
(i.e. <br />
u_n <br />
diverges) then <br />
\sum\limits_{n = 1}^\infty  {( - 1)^{n + 1} } u_n <br />
diverges?
    yes
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  5. #5
    Member RedBarchetta's Avatar
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    Ummm, Paul's Calculus page says the limit going to zero is one of the conditions...

    Pauls Online Notes : Calculus II - Alternating Series Test
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    Ummm, Paul's Calculus page says the limit going to zero is one of the conditions...

    Pauls Online Notes : Calculus II - Alternating Series Test
    it is a condition for convergence, not a condition to use the test

    to use the test, we must have that the u_n's are non-nonnegative, then if they are decreasing and we also have the limit going to zero, we have convergence
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