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Did I correctly prove the divergence of this series?

  1. Jul 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Verify that the infinite series diverges.
    I have the series from n=1 to infinity of (2^(n)+1/2^(n+1)


    2. Relevant equations
    Nth term test(This is the way the book did it but I did it used the geometric series test
    and I just want to verify if my Algebra was correct)


    3. The attempt at a solution
    First I split the series into two separate series and let the series go from n=0 to infinity.
    So I have 2^(n+1)/2^(n+2) add 1/2^(n+2).
    I believe I can then bring down the some of the exponents and simplify to
    2*2^n/4*2^n add 1/4*2^n
    so for the first series I let a=1/2 and r=1^n
    because |r|=|1| is greater than or equal to 1, the series diverges.
    I completely ignored the second series that I made because its irrelevant to simplify
    as I already know the first series diverges.

    Is this proof valid and was my algebra correct?
     
  2. jcsd
  3. Jul 13, 2012 #2

    Mark44

    Staff: Mentor

    Is this your series?

    $$ \sum_{n = 1}^{\infty} \left(2^n + \frac{1}{2^{n+1}}\right)$$

    You are missing a right paren, so I'm not sure what you intended.

    The easiest approach is the n-th term test for divergence. It seems like a lot of extra work to try to make this series look like a geometric series, for which the general term is arn.
     
  4. Jul 13, 2012 #3
    (2^(n)+1)/2^(n+1)

    I apologize for the confusion, that is my series.
    I agree that the easiest approach is the n-th term test, however I don't always see the easiest approach every single time.
    The problem with series is that you can do the wrong thing and still get the right answer, I just want to make sure my proof was right
    since I got the right answer.
     
  5. Jul 13, 2012 #4

    micromass

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    I think that if you're given a series, that the first thing you should always try is the n-th term test. If you don't find it using that, then you can move on. But the n-th term test is usually a lot easier than the other tests.
     
  6. Jul 13, 2012 #5
    Thank you for the advice, I will always do that from now on. I kinda saw that I should've done that halfway through the problem but I thought to myself why not finish it.

    Could anyone though please confirm if my Algebra was correct? I have a test coming up and there will be geometric series that require this kind of algebraic manipulation and I just want to know if everything I did algebraically was correct.
    I mostly just need to know if the exponent rule I did was right.
    Like when you have 2^(n+2), its actually 2^2 * 2^n. This is a valid right?
     
  7. Jul 13, 2012 #6

    Mark44

    Staff: Mentor

    Like so:
    So this is your series.
    $$ \sum_{n = 1}^{\infty} \left( \frac{2^n + 1}{2^{n+1}} \right)$$
    But you should try the easy approaches first, and the n-th term test for divergence is probably the easiest of all the tests.
    Here's from your first post...
    Wouldn't it have been simpler to leave the index starting from 1?
    Then you would have 2^n/(2^(n + 1)) + 1/(2^(n + 1)).
    Use + instead of "add".
    2^n/(2^(n + 1)) + 1/(2^(n + 1)) = 1/2 + 1/(2^(n + 1)).

    The first part of what you have is correct, but can be greatly simplified.
    2*2^n/4*2^n = 1/2


    No. For a geometric series r has to be a constant. Of course, for any finite n, 1^n is 1, so was this a typo?

    Your first series is $$ \sum_{n = 1}^{\infty} \frac{1}{2} $$
    This is not a geometric series.

     
  8. Jul 13, 2012 #7

    micromass

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    Your algebra was perfectly fine.

    But there was a mistake in the proof: you split the series in two parts

    [tex]\sum{a_n+b_n}=\sum a_n +\sum b_n[/tex]

    and deduce from that that a series in the right hand side diverges, thus the left-hand side diverges as well. This is not true. For example

    [tex]\sum{1-1}=\sum 1 - \sum 1[/tex]

    the two series in the right-hand side diverge, but the left-hand side does not diverge.

    The rule

    [tex]\sum{a_n+b_n}=\sum a_n +\sum b_n[/tex]

    only holds if both [itex]\sum a_n[/itex] as [itex]\sum b_n[/itex] converge. If both diverge, then there is nothing that can be said.
     
  9. Jul 13, 2012 #8
    Thank you guys so much for the thorough explanations! I really need to review my property of series. My biggest concern was the algebra but thankfully I did that right.
    As for the use of add, I thought it would be easier to read if I used add instead of +(not sure why).
     
  10. Jul 13, 2012 #9

    HallsofIvy

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    Well, if [itex]a_n> 0[/itex], [itex]b_n> 0[/itex], and that is the situation here, and either [itex]\sum a_n[/itex], or [itex]\sum b_n[/itex], or both diverge, then [itex]\sum (a_n+ b_n)[/itex] diverges.
     
  11. Jul 13, 2012 #10

    Mark44

    Staff: Mentor

    One point that micromass made was very pertinent - about splitting series into parts and when you can do that. That thought came to mind while I was writing a response, but I didn't include it, so I'm glad he did.
     
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