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Homework Help: Are my answers right? (Calculus! YAY )

  1. Nov 12, 2004 #1
    Are my answers right? (Calculus! YAY!!!)

    I need to know if my answers to these questions are correct... Any help is very much appreciated and thanks in advance. :)

    Questions = bold

    Does [tex]\sum \frac {ln k}{\sqrt k}[/tex] converge or diverge? limits are 0 to infinity.

    answer: diverge because the limit of [tex]a_n[/tex] goes to 1 and not zero.

    Does [tex]\sum \frac {cos^2 k}{k^2 +1}[/tex] converge or diverge? Limits are 0 to infinity.

    answer: converge because it's smaller then [tex]\sum \frac {1}{k^2}[/tex] which converges because of the p-series test... I'm not sure about this since k is inside cosine... :(

    What are the radius and domain of this power series: [tex]\sum \frac {n(x-3)^n}{2^(n+1)}[/tex] limits are 0 to infinity.

    answer: Radius is 2 and domain is [ 1, 5)...

    What I did here was the ratio test which ended up with a limit of [tex]\frac {1}{2}[/tex] and then got to x being between 1 and 5. I checked both points and the series diverges with 5 plugged in for x and converges (alternate series) when 1 is plugged in...

    whew. Again, thanks for any help. This is pretty important right now.
  2. jcsd
  3. Nov 12, 2004 #2


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    How did you get that the limit of the terms is not zero? You might want to check this.

    That's fine, you can compare the absolute values of the terms in your series to [tex]\frac{1}{k^{2}}[/tex] and conclude your series converges (absolutely). Remember the absolute value of the cos part is no greater than 1.

    Something went wrong with your latex I think, is your series:
    [tex]\sum \frac {n(x-3)^n}{2^{n+1}}[/tex]

    When you put 1 in for x, what's your series? Can you state the conditions of the alternating series test?
    Last edited: Nov 12, 2004
  4. Nov 12, 2004 #3
    Ugh, I knew I didn't get those right. :frown:

    I just thought that both lnx and x^1/2 go to infinity slowly but they do end up getting bigger and bigger and in the end it's infinity over infinity which I guess is one... but I guessed wrong, right? :(

    nice to hear I do have a brain... not a good one though. :p
    Yeah that's what I meant.
    conditions for alternating series:

    1) limit [tex]a_n[/tex] goes to zero
    2) absolute value of [tex]a_n[/tex] is bigger then [tex]a_(n+1)[/tex] and the same for the values after that.
    2) [tex]a_n > 0[/tex] and [tex]a_n+1 < 0[/tex] and it goes back to bigger then zero for n + 2 and then smaller for the one after that and so on...

    When I plug in 1 I get:

    [tex] \sum \frac {n(-2)^n}{2^n 2^}[/tex]

    Now after that I don't really know what to do since the top 2 has a negative I'm not sure I can cancel it with the bottom... so I just assumed it's an alternating series. :redface:
    Last edited: Nov 12, 2004
  5. Nov 12, 2004 #4


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    x and x^1/2 both go to infinity, but x/(x^1/2) clearly doesn't go to 1.

    I'm not sure you have the tools yet to deal with the limit of the terms of this series, but you don't need to in any case here (it won't actually help).

    Try a simple comparison test, remembering that ln(x)>=1 if x is large enough.

    You can cancel, and you can break up the numerator as [tex](-2)^n=(-1)^{n}2^{n}[/tex] then cancel. It's definitely an alternating series from the [tex](-1)^n[/tex] term, you are right about that (this is your condition 3)). But you can't conclude from the alternating series test that it converges unless you've verified 1) and 2) as well. Do the cancelling, then what do you say about these conditions?
  6. Nov 12, 2004 #5
    For the first one, just because both terms go to infinity, this does not necessarily mean that the limit is one. Equations of the form [tex]\frac{0}{0}[/tex] or [tex]\frac{\infty}{\infty}[/tex] are known as the indeterminate case. They may both approach the same number, but one number may do it much faster than the other, for instance in the fraction [tex]\frac{x}{x^4}[/tex] both numbers approach [tex]\infty[/tex] as x approaches [tex]\infty[/tex], however x^4 does so much faster that the actual fraction overal approaches 0. A good way to find the limit of a function as it approaches infinity is to use L'Hopitals Rule, basically take the dirivative of both the numerator and the denominator independently, and then evaluate those at the limit points. Using this rule you can find how the original function converges using the dirivatives of both parts of the fraction.
  7. Nov 13, 2004 #6
    I'm sorry but I don't agree with you about L'Hopital's Theorem as the subsequent computation of derivatives can not only be tedious but misleading as in the case of

    [tex]\lim_{x\rightarrow 0} \frac{\sin(\sqrt{x})}{\sqrt{x}}[/tex]

    which CANNOT be computed using L'Hopital's Rule simply because the function is not differentiable at x = 0. However a mere substitution, viz. [tex]y=\sqrt{x}[/tex] gives the answer. Same goes for several rational functions which through a bit of thought can be reduced to previously known limits (some of which may have geometric significance).

    Another example is

    [tex]f(x) = \frac{\tan(x)}{\sec(x)}[/tex]

    and we are required to find the limit of f(x) as x tends to 0. Clearly L'Hopital's Rule is a bad idea...a simpler way is to express the numerator and denominator (in the limit) in terms of sin(x) and cos(x) so the problem reduces to finding the limit of sin(x) as x tends to 0.

    I'd suggest that you avoid overuse of the theorem and resort to it only when the function appears to be irreducible or when nothing else comes to mind. Often, a geometric or order of magnitude argument such as the one used by Lyuokdea is a good idea too.

    Hope that helps...

    Last edited: Nov 13, 2004
  8. Nov 13, 2004 #7

    [tex] \sum \frac {n(-1)^n}{ 2^}[/tex]

    I'm guessing this goes to infinity and doesn't satisfy condition 1 for alternating series. :(

    Thanks guys for all the help. I learned a lot from this. :)
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