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Series convergence tests

  1. Apr 17, 2014 #1
    I am currently learning series and testing for convergence. For comparison tests especially I am having an issue grasping the concept of picking a proper limit to compare too.

    For example the following problem

    If someone could please put it in the form where it actually looks like what it does on paper that would be great. I am not quite sure how to do that yet.


    In the solutions my teacher provided it states to use the comparison test with 1/n^2 and since that is convergent than the original being less all the time is also convergent. I understand that part, I just don't get how to figure out how 1/n^2 was chosen.

    Another one is limit comparison for sin(1/n)/n. This one I am clueless about and do not understand the solution at all.

    The solutions state to compare to 1/n^2 and then it states lim n→∞ (sin(1/n)/(n(1/n^2)) and then that becomes lim n→∞ an/bn = 1/n^2/n^2 which is n^2/n^2 = 1 and that would mean its convergent.

    I know how to tests work but just not how these comparison limits are chosen and especially for the second one the solution is not clear how to get from one step to the other.

    Thanks in advance.
  2. jcsd
  3. Apr 17, 2014 #2


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    You want to start approximating how large quantities are for large n. If n is very large, then n^2-1 is approximately n^2 so sqrt(n^2-1) is approximately sqrt(n^2)=n. (n^3+2n^2+5) is approximately n^3, since 2n^2+5 is much smaller than n^3. So you are left with n/n^3=1/n^2. For the second one, sin(x) is approximately x if x is small. Since for n large, 1/n is small, sin(1/n) is approximately 1/n. Etc. Beyond that you need to make a better argument to actually show convergence or divergence. But that's the first step in guessing a comparison.
  4. Apr 17, 2014 #3


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    When you have polynomials, the dominant term is the highest power and the other terms are small in comparison as n gets large. So for large n, you would think ##\sqrt{n^2-1}## is "like" or approximately ##\sqrt{n^2}=n##. Try it for a few large ##n## and you will see what I mean. Similarly the denominator is dominated by the ##n^3## so you would think the whole fraction would grow like ##\frac{n}{n^3}=\frac 1 {n^2}##.

    Again, as ##n## gets large, ##\frac 1 n## nears 0. Recall that ##\lim_{x\to 0}\frac {\sin x} x = 1## so ##\sin x \approx x## for small ##x##. So for large ##n##, ##\sin(\frac 1 n) \approx \frac 1 n##.

    [Edit] Dick types faster...
  5. Apr 17, 2014 #4


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    Lets start with the first one.


    Notice the higher order terms dominate the expressions in both the numerator and denominator. That is, as n gets arbitrarily large, the higher order terms are 'more significant' than the lower order terms and the constant terms. So you could write something such as:

    $$\frac{\sqrt{n^2-1}}{n^3+2n^2+5} ≤ \frac{\sqrt{n^2}}{n^3} = \frac{1}{n^2} $$
  6. Apr 17, 2014 #5
    Okay great, I am set on the first one now. Completely understand. Thanks
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