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  1. Default Proving the convegence of the series sqrt(n)/(n^2 - 3)


    I have a series sqrt(n)/(n^2 - 3) that goes from 2 to infinity, and I want to see if it converges or diverges. I tried the ratio test, the test for divergence, and the comparison test, but none of them work (I didn't use the integral test because the teacher will not have a problem where the integral is complex for most people to do). Is there any way for me to prove that the series converges or diverges?

  2. Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    Why doesn't the ratio test work?

  3. Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    You end up with 1.

  4. Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    My bad. Brain fart.

    You should be using the limit comparison test:

    which states lim(n->infinity) a_n/b_n = c, 0 < c < infinity
    where a_n and b_n >= 0;

    the series a_n converges if b_n converges

    For very large n, you can approximate
    sqrt(n)/(n^2-3) ~= sqrt(n)/(n^2) = 1/n^3/2 = b_n
    (as n approaches infinity, -3's effect approaches 0)



    Since

    converges, (p-series p = 3/2)

    therefore

    converges

    These images work, right?

  5. Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    You only have so many tools for these problems. There was only one problem that I had to get creative with, and it involved a limit comparison test. I still have no idea how I thought of that b_n, but it turned out to work.

  6. Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    If you don't want to be like "for very large n", you can also observe that
    sqrt(n)/(n^2-3) < sqrt(n)/(n^2/2)
    for n>2 this holds, so you can squeeze the terms under 1/2*n^(-3/2) which is half a convergent p-series.
    n=1 and n=2 are not a problem cause there are only a finite number of them, and none are infinite.


    It'd be simpler on something like sqrt(n)/(n^2+2) because you can directly observe
    sqrt(n)/(n^2+2) < sqrt(n)/(n^2)
    for any n>0.

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    Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    What level math is this? I don't think I've ever seen this before.

  8. Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    Limits are usually introduced before calculus, as understanding them is part of the calculus fundamentals. So it could be 12th grade or first year university, where I live.

    How detailed it gets probably depends on the audience of the course, should at least see them by calc though.

  9. Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    Calculus C/first year college math for most people.

  10. Default Re: Proving the convegence of the series sqrt(n)/(n^2 - 3)


    Not all first years take Calculus 2, most don't even get to that level. Some will stop at Calculus 1, others begin with college algebra. I think you give people too much credit in their mathematical "prowess". It's hard to find people who enjoy math enough to take it to a higher level besides those that are required such as engineering.

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