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Direct and limit comparison test?

  1. Aug 5, 2008 #1
    Does anyone have any knowledge on these? I look at an example of a direct or limit comparison and i see what they are doing, but I have no idea how they got the "comprable" term.... for example the sum (n = 1) to infinity n/(n^2 +1) you can use either test, but how do you choose what to compare it to? this is my issue and it occurs on other problems where you have to use comparison... I know for direct comparison you need to compare the original sum to something smaller or larger than itself and depending on what it does will tell you if it converges or diverges. I know this, but I do not know what to compare it to!! Do you just pick something out of thin air that is in the same general form as the sum and make it smaller or larger? aaaarrrghhhhh!!!!!
  2. jcsd
  3. Aug 5, 2008 #2
    Direct comparison:
    Let's say we want to show that [tex]\sum_n a_n[/tex] converges by direct comparison. Basically, we use trial-and-error to find a series [tex]\sum_n b_n[/tex] such that for all n greater than some finite value M, [tex]b_n \geq a_n \geq 0[/tex]. Think of it as the sequence [tex]b_n[/tex] squeezing [tex]a_n[/tex] into the x-axis.
  4. Aug 5, 2008 #3
    that made it clearer for that test, thank you, but I am still unsure about the limit test. It seems to be very similar
  5. Aug 5, 2008 #4
    Well experience certainly helps. But I think the best way is to consider the behavior of the function for large values of n (for instance take a look at the limit as n -> +inf). Sometimes you do have to try an algebraic manipulation (often dividing top and bottom by some factor).

    But really I would consider large values of the n and remember that for direct and limit comparison, you are often comparing to a geometric series or p-series. Taking your example as an example.

    n/(n^2 +1) behaves like 1/n for large values of n. Then the sum behaves like a divergent p-series with p = 1 (very common, and remember p-series is divergent is p <= 1). Now consider the comparison test. It looks like it wouldn't help too much because our original series seems to be smaller than 1/n for all n (why would direct comparison fail in this case?).

    Now try limit comparison. Find the limit [n/(n^2 + 1)]/[1/n] as n -> +inf. Can you now determine convergence/divergence of that series?
  6. Aug 5, 2008 #5
    Remember for limit comparison, we take the expressions in the two series we are comparing and form the ratio. If the limit L of that ratio as n -> +inf exists and is greater than 0, then the test is conclusive. If we know one of the series was convergent, then the other converges as well. Similarly, if one was divergent, both are divergent.

    Just remember that it often comes down to comparing a given series to a geometric series or a p-series when we're using direct/limit comparison.

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