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Alternating Series Help

  1. Nov 21, 2005 #1
    Alternating Series Help!!!

    I apologize right now for the fact that I have no idea how to use LaTeX

    I can't figure out if the following alternating series is convergent or not:

    Sum(((-1)^(n-1)) * ((2n+1)/(n+2))) from 1 to infinity

    the root test is not applicable, A(n+1)>An, and the ratio test gives me Limit=1, so I have no comclusive evidence either way. Even Maple 10 couldnt give me an answer.
    I have a test tomorrow. HELP!!!!!!!
  2. jcsd
  3. Nov 21, 2005 #2
    You just need to use the alternating series test

    the series \Sum_{n=1}^{\inftty} (-1)^n a_n converges if

    lim_{n->\infty} a_n =0 & a_n is monotonic decreasing
    Last edited: Nov 21, 2005
  4. Nov 21, 2005 #3
    Hey how can you tell if the series in monatonic decreasing?

    And thank you very much.
  5. Nov 21, 2005 #4
    That would be great if lim an=0 (n->inf) but it doesn't. lim an=2. Although the first criterion is met for the alt. series test, the second is not. In fact all tests about regular convergence fail on this series. If you are allowed to broaden your idea of convergence it may be Abel or Cesaro summable. Otherwise, every other test for "regular" convergence is inconclusive. Unless there is a trick I am missing you might not know whether it converges in traditional sense.
  6. Nov 21, 2005 #5
    lol, it just killed Matlab 7.0 too! "NaN"
  7. Nov 21, 2005 #6
    If {a_n} is a (strictly) monotonically decreasing series then

    (a_(n+1))/a_n < 1 . (for all n)

    For the alternating test both this and the limit must hold,
    hence if one fails, the series does not converge
  8. Nov 21, 2005 #7
    the Lim=2 as n->infinity

    Does this mean that it is definitely divergent then, since neither criteria for convergence was met?
  9. Nov 21, 2005 #8
    mathphys, are we going to argue all night. I will state the alternating series test: An alternating series sum((-1)^n*an) converges IF: an is monotone decreasing AND lim(an)=0. We have the prepositional logic statement: (A+B)-->C. Now you are saying that if we negate either A or B, ~A or ~B, then that means ~C? WLOG, say we have A and ~B. Then we have ~(A+B). So your claim is that given (A+B)->C and ~(A+B), then we may conclude ~C. That is a logical fallacy called denying the antecedent. Therefore it is still unknown whether the series converges or diverges.
    Last edited: Nov 21, 2005
  10. Nov 21, 2005 #9
    Hate to be a burden here, but I have 2 Power series questions as well

    I have to find the radius of convergence for each of these:




    Im sure I could work them out, but the test is in 4 hours and my brain ceases functioning at this hour.

    Thanks for your help so far, guys.
  11. Nov 21, 2005 #10
    "For the alternating test both this and the limit must hold,
    hence if one fails, the series does not converge", i wrote, "hence if that one...", I must had written ;).... meaning that if the limit criteria is not satisfied the series diverge.

    Lim_{n-->infinity} a_n=0 is a necessary condition for convergence of the alternating series. If this criteria fails the series diverge. Nevertheless, if this requierement is satisfied this does not mean that the series converge, because is necessary but not sufficient.

    However, in this example this limit does not hold, so the series does not converge (for THIS example).
    Last edited: Nov 21, 2005
  12. Nov 21, 2005 #11
    For the first one: B=lim(sup(abs(an)^(1/n),N),N,inf)=1. Therefore 1/B=1=Radius of convergence. For the other the test fails, so give me a moment.
  13. Nov 21, 2005 #12
    Using the ratio test on the second one, lim(abs(a(n)/a(n+1))=1/2. Therefore R=1/2.
  14. Nov 21, 2005 #13
    mathphys, you fail to understand the logic behind conditional statements. The only inferences that may be made using the conditional statement A->B, are Modus Pollens and Modus Tollens. In the former, we have A->B, A, therefore B. In the latter A->B, ~B, therefore ~A. We have neither of the two cases in this situation and by continuing to support the claim that since lim an=2 then the series diverges, shows that you are unwilling to put as much effort in understanding prepositional logic as you are to argue with people on the internet. You are commiting a logical fallacy, denying the antecedent, by saying for certain it converges. Example: "If I was killed in a plane crash, then I am dead." I was not killed in a plane crash. By your logic, we may infer that I am not dead? Does that make any sense?
  15. Nov 21, 2005 #14

    I won't discuss logic with you, because this is not the place or post, nor i will put simplistic non sense examples.

    Just try understand what necessary and sufficient conditions mean.
  16. Nov 21, 2005 #15
    My friend, you are wrong. I gave a proof of why you are wrong. (A+B)->C, ~(A+B), therefore ~C is INCORRECT, denying the antecedent. Perhaps I didn't make clear what A, B, and C are. A=an is monotonically decreasing, B=lim(an)=0, and C=an converges. Now we know A to be true, we know B to be false, therefore ~B. Now, in order for A+B to be true, BOTH A and B must be true, therefore we have ~(A+B). It is therefore INCONCLUSIVE whether we have C or ~C. Give me a PROOF otherwise? You claim that I don't know necessary and sufficient conditions, give me a proof of why I am wrong. And by the way, my example wasn't stupid. just cuz I didn't get killed in a plane crash, doesn't mean that I am not dead, I could have died by other means. Perhaps going insane on the internet why attmepting to persuade the thick headed.
  17. Nov 21, 2005 #16
    ....i didnt mean to start a fight
    i just wanted homework help
    feel free to continue though, im not really knowledgeable on logic and proofs, so maybe i can learn something from this debate

    Thanks again for your help guys
  18. Nov 21, 2005 #17

    There is no point in that , it won't help you with your homework, nor me. :)
    Just for reference
    http://planetmath.org/encyclopedia/AlternatingSeriesTest.html [Broken] see this and references therein.


    "To say that A is necessary for B is to say that B cannot be true unless A is true, or that whenever (wherever, etc.) B is true, so is A."

    Because the limit condition ALONE (not A+B nor nothing like that :P) is a necessary condition for the alternating series to converge, it cannot be true that the series converge unless the limit condition (ALONE!) is satified. Because it is not, is not true that the series converge. Then what follows? you guess it :).
    Last edited by a moderator: May 2, 2017
  19. Nov 21, 2005 #18
    wait, for the second power series, after some manipulation, wouldnt you end up with 2x* Lim((1+1/n)/(1+2/n))^(n+1), which would be 2x/e, and the radius would be e/2?
    Last edited: Nov 21, 2005
  20. Nov 21, 2005 #19

    You're on the right track, an exponential has to appear, just work it a little more
    Last edited: Nov 21, 2005
  21. Nov 21, 2005 #20
    well (1+1/n)^(n+1) is e
    and 1 + 2/n)^(n+1) is e^2
    so it would have to be 2x*e^-1, or 2x/e, right?
    or am i missing something
    2 hours till test............
    i love caffeine
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