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Comparison Tests of Convergence 1 + 1/1! + 1/2! + 1/3! + · · ·?

  1. Aug 15, 2010 #1
    1. The problem statement, all variables and given/known data
    question given is 1 + 1/1! + 1/2! + 1/3! + · · ·
    the question need to be compared with another equation
    so how to get the second equation?


    2. Relevant equations
    n/a


    3. The attempt at a solution
    i duno hw to startT.T
     
  2. jcsd
  3. Aug 15, 2010 #2
    well, first off you know that if one series' terms are less than another's that converges, then they both converge.

    if youre limited to using the comparison test, just try to look for a series with terms greater than 1/n! as n tends to infinity.

    examples might include a p-series of some sort: ( 1/(n^p)) where p>1. although i think that factorials (n!) always surpass all exponentials (x^n) (side note: does anyone know of a proof or theorem for that?) so you might wish to try for a different convergent series.
     
  4. Aug 15, 2010 #3
    is tat means there are few different equation can be use to compared tis equation?
     
  5. Aug 15, 2010 #4
    sort of. with these kinds of problems that ask you to use the comparison tests, you cant rely solely on one way of doing it. you have to be creative, because there are probably tons of different solutions.

    you're trying to prove that the series (1/n!) converges, right?
     
  6. Aug 15, 2010 #5
    ya,and the answer given is 1st equation compare with 1+1+1/2+1/2^2+1/2^3+...
    how to noe we shud use tis equation to compare with the 1st equation?
     
  7. Aug 15, 2010 #6
    well, the first thing youll do with that is compare them.

    you'll do this by making an inequality (> or <) with the two series. im assuming you know (or they told you) that 1+1/2+1/2^2+1/2^3+... converges, right?

    so once you compare them to see which one is greater than the other, you can draw your conclusion as to if it converges or not.
     
  8. Aug 15, 2010 #7
    but the problem is in the question din give me the 2nd equation,so how do we determine the 2nd equation based on only the 1st equation?
     
  9. Aug 15, 2010 #8
    the idea behind convergence tests is very theoretical. you are not trying to "make" the "equations" look like eachother. the idea is to prove that the series we dont know about is less than the series we do know converges, so that we can say that it in fact does converge. does that make sense? or am I hitting on something kinda different?
     
  10. Aug 15, 2010 #9
    ok,nw i try to make example,u c whether i'm getting ur meaning right onot:)
    nw,question given is 1 + 1/1! + 1/2! + 1/3! + · · ·
    tis is the 1st equation

    if i use 1st equation to compare with 1+1/2+1/2^2+1/2^3+... (which is converges) and 1st equation is less than 2nd equation,thn the 1st equation is converge

    but
    if i use the 1st equation to compare with another equation which is also converges but tis time the result is 2nd equation is less than the 1st equation,thn the 1st equation is diverge?

    am i right?0.0
     
  11. Aug 15, 2010 #10
    it seems that you are starting to understand it better.

    your first example is correct. if the series that converges is greater than the one in question, it converges.

    however, your second example is incorrect. if we compare the series in question (the "1st equation") to a series that converges, but it turns out that the 1st equation is greater than the converging one, it doesnt necessarily mean that it converges OR diverges.
    think about both alternatives. you could have the "1st equation" still converge AND be greater than another converging series. or on the other hand, you could have the "1st equation" be a DIVERGENT series and STILL be greater than the convergent series. The problem is that we cannot determine what its behavior is with this certain series. another one must be found that can show it converges (i.e. one that converges itself.)

    so no, if a series is greater than another convergent series, it does not necessarily mean it diverges.

    however, if you showed that a series (lets stick with "1st equation") is greater than a DIVERGENT series, then you know for a fact that it diverges. for example, if we say that series A is divergent, then any series BIGGER than a divergent series could not ever converge.
     
  12. Aug 15, 2010 #11
    k,i get wat u mean:)
    so wat about the 2nd equation,we jz use any equation we lik as 2nd equation?
     
  13. Aug 15, 2010 #12
    wat u mean by 'one that converges itself'?
     
  14. Aug 15, 2010 #13
    Yes. The idea is to keep testing different series in hopes of finding a series that converges that is greater than the series in question.

    And by that I just mean you need to keep testing series that converge.
     
  15. Aug 15, 2010 #14
    ic,thx for ur patient^^
    few more thng tat bother me

    i being searching for some reference on net and i found that in chapter comparison test,the example given is the series A is compared with series B while series B is change(expand/transform/intergrate/simplify/or watever)from series A.y is tis happening?
    isn't tis different frm wat u teaching me?(keep testing the series)
    tis is the webpage
    http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx
    i'm not saying tat u teach me wrong,jz vr blur:)
     
  16. Aug 15, 2010 #15
    A lot of what he does there is somewhat unnecessary. what he did was show that the series in question was less than another series by comparing each one's partial sums. if youre just looking to see whether or not a series converges, you dont need to go through all of the proofy partial sum stuff unless you are asked to. as with the integral, may i ask if you know about any other convergence theorems? all hes doing with that is showing how another convergence test, the integral test, doesnt work all of the time.
     
  17. Aug 16, 2010 #16
    actually i duno anythng about convergence,i jz help my fren to ask question bcoz she going to exam soon and she cant on9 to search for info^^sryXP
     
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