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Series and limit help for analysis

  1. Nov 2, 2006 #1
    I need help proving that if (xn) is a convergent series, then (xn)^2 is a convergent series.

    also is (n^3)*(xn) is convergent, then (xn) is convergent. i reasoned that if (n^3)*(xn) is convergent, we know that (n^3)*(xn) > (xn) > 0, therefore by comparison test, (xn) is convergent.

    determine whether or not the function exists and if it doesnt explain why:

    lim x--> 0 (x^2/(|x|))
    i cant put it in formal terms why this limit does not exist (we did not go over continiuty yet)

    show that the function
    f(x) = { x if x is rational, 0 if x is irrational
    has a limit at p=0 and does not have a limit at any other point.

    cant quite establish that the limit exists and how do i prove that no other points have limits?????
     
  2. jcsd
  3. Nov 2, 2006 #2

    StatusX

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    You can use the comparison test here.

    Looks good.

    What makes you think it doesn't exist? What does the function look like?

    For a limit to exist, it must be the same no matter how you approach the point. So if you can find two sequences x_n and y_n with both x_n->x and y_n->x, but f(x_n)->a and f(y_n)->b, then the limit at x doesn't exist.
     
    Last edited: Nov 2, 2006
  4. Nov 2, 2006 #3
    how could i use comparison test for the first one????? sum(xn) is convergent, but sum(xn)^2 is greater than sum(xn) and therefore comparison test is not applicable, you need the greater series to be convergent to deduce that the lesser series is convergent
     
  5. Nov 3, 2006 #4
    bump any help????????????

    i still do not get how to solve the sum(an) converges then sum(an)^2 converges, prove or disprove

    i still don't get how lim x--> 0 (x^2/(|x|)) has a limit or doesnt have a limit

    or how to show that f(x) = {x if x is rational, 0 if x is irrational) has only a limit at p = 0 and there are not limits at any other point.

    big help is needed, im so lost
     
    Last edited: Nov 3, 2006
  6. Nov 3, 2006 #5

    quasar987

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    (a_n)²is not necessarily greater than a_n. Take a_n=½ for instance. Then (a_n)²=¼<½. And it just so happen that when a series converges, the limit of its "argument" is 0, so for sufficiently large n, a_n<0 ==>a_n^2<a_n. So by the comparison test, (a_n)² converges.

    You can also do it by the ratio tets. The esssence of the ratio test is that given two series [itex]\sum a_n[/itex], [itex]\sum b_n[/itex] for which you know that [itex]\sum a_n[/itex] converges, consider the limit of the ratio of their general term:

    [tex]\frac{b_n}{a_n}[/tex]

    and you know that since [itex]\sum a_n[/itex] converges, then [itex]a_n\rightarrow 0[/itex], so if the limit of the ratio is 0, then it means that b_n goes to 0 faster than a_n. Then surely, [itex]\sum b_n[/itex] will converge (more rigorously, it is so by the comparison test for sequences applies to the sequences of partial sums). Also, if the ratio "stabilizes" at infinity, i.e. equals some constant L, then b_n dives as fast as a_n so [itex]\sum b_n[/itex] must converge as well.

    I don't know why I'm telling you this; it's for me I guess. But the point is that it is obvious that by taking a_n as your series and b_n = (a_n)², then it works.
     
    Last edited: Nov 3, 2006
  7. Nov 3, 2006 #6

    quasar987

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    Would it help you see it if you wrote x² as |x|²?

    What do you know about the density of the rationals and inrationals is R?

    Use that and the fact that if the limit exists, it is unique.
     
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