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Calculus II - Infinite Series - Integral Test Problem

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine whether the following series converges absolutely, converges conditionally or diverges. Show your work in applying any tests used. sigma[k=1,inf] [(-1)^k*k/sqrt(k^4+2)]

    2. Relevant equations

    integral csc(x) dx = -ln|sec(x)+cot(x)| + C
    csc(x) = 1/sin(x)
    sec(x) = 1/cos(x)
    cot(x)= 1/tan(x)
    tan(x) = opposite/adjacent
    sin(x) = opposite/hypotenuse
    cos(x) = adjacent/hypotenuse
    d[cot(x)]/dx=-csc(x)^2
    x^a*x^b = x^(a+b)
    x^a/x^b = x^(a-b)

    3. The attempt at a solution

    I tried using the divergence test and applied Lpitals rule (or however you spell it) multiple times and just gave up so I then went on to tried to proceed using the integral test.

    My issue came when I was trying to take the integral of k/sqrt(k^4+2). I got -1/2 ln|sqrt(k^4+2)/sqrt(2) + k^2/sqrt(2)|.
    I don't see what I'm doing wrong. I took the derivative of this using wolfram alpha, http://www.wolframalpha.com/input/?i=d%5B-1%2F2ln|sqrt%28k^4%2B2%29%2Fsqrt%282%29%2Bk^2%2Fsqrt%282%29|%5D%2Fdk and got -k/sqrt(k^4+2), which isn't equal to my original integrand, k/sqrt(k^4+2). Apparently the correct antiderivative of the integral is 1/2ln|sqrt(k^4+2)/sqrt(2)+k^2/sqrt(2)|, as you can see by checking with wolfram alpha http://www.wolframalpha.com/input/?i=d%5B1%2F2ln|sqrt%28k^4%2B2%29%2Fsqrt%282%29%2Bk^2%2Fsqrt%282%29|%5D%2Fdk, it's correct.

    So apparently I'm not suppose to have a negative in front of my antiderivative but I don't see were I went wrong, I guess I was suppose to have -1/2 integral csc(theta) dtheta, which would of given me the correct antiderivative but I don't see why I'm suppose to have a negative sign out there, when I substituted k and dk the +/- and +/- simplified to just +, I think, which doesn't need to be shown because something is generally assumed to be positive if there's not a negative sign in front of. This however is the only place that I can think I made a error, in simplifying +/- * +/-, as this is the only point in the evaluation that I messed around with positive and negative signs. This leads me to believe that +/-*+/- can be simplified to - and not +, but this is incorrect, but I don't see were else I could of gone wrong with positive and negative signs.

    I don't see what I'm doing wrong. Thanks for any help! Let me know if you can't follow my work, apparently it's all right except for the negative symbol in my antiderivative that's not suppose to be there. See the attachment. Click on it to view it in a window that pops up. If you want to view it at a larger scale click on it again to view it in a new tab blown up some.
     

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  3. Sep 4, 2011 #2

    dynamicsolo

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    I am reading this as the series [itex]\sum_{k = 1}^{\infty} \frac{(-1)^{k} k }{\sqrt{ k^{4} + 2 } }[/itex]. The Integral Test should probably be used as a last resort, since it's incredibly easy to create difficult to impossible integrals.

    When making a comparison test, you should drop the "alternating-sign factor" and find a way to eliminate the radical. Since [itex]\sqrt{ k^{4} + 2 } > k^{2}[/itex], it will be the case that [itex] \frac{k}{\sqrt{ k^{4} + 2 } } < \frac{k}{k^{2} }[/itex] ; this will lead to a comparison with a familiar series which diverges.

    Unfortunately, the comparison runs the "wrong way" for divergence: knowing that a series gives a sum smaller than a divergent series tells us nothing about convergence. So we will need the Limit Comparison Test; our previous attempt gives us a clue as to what to try. We use our general term, again without the [itex](-1)^{k}[/itex] factor, and compare it to 1/k . One has to exert a little care as to which way to set up the ratio; we will use

    [tex]\lim_{k \rightarrow \infty} \frac{\frac{1}{k}}{\frac{k}{\sqrt{ k^{4} + 2}}} .[/tex]

    Once you sort out the algebra, you will have a limit you can evaluate using division in the numerator and denominator by k2 . (Don't try to use L'Hopital's Rule: radicals are its Achilles Heel and are impossible to be rid of by differentiation!) The result of this test will tell you that the series is not absolutely convergent.

    You can then try the Alternating Series Test from there.
     
    Last edited: Sep 4, 2011
  4. Sep 4, 2011 #3
    The latex or whatever it's called didn't seem to work in some of that post, when I try it in wolfram alpha http://www.wolframalpha.com/input/?i=sum+k%2Fsqrt%28k^4%2B2%29%2C+k%3D1+to+inf, it says that
    The ratio test is inconclusive.
    The root test is inconclusive
    By the integral test, the series diverges.

    I thought the integral would be easy to evaluate, I almost had it as well, still don't see were I went wrong, kind of would like to know were I went wrong in evaluating the integral just for the thrill of it because it doesn't seem as if I went wrong any where as far as I can see.

    Unfortunately I dislike using the comparison test and limit comparison test and shy away from it. I'm not that good at coming up with series to compare with. I need to become more stronger in using it I guess.

    I never thought of applying the alternating series test. I'm kind of lost as to how I would because I would have to take the limit as k->inf

    nevermind I can see it now lol for somereason it shows up and sometimes and dosen't
     
  5. Sep 4, 2011 #4

    dynamicsolo

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    The Root Test is flat useless here; the Ratio Test is likewise, since the general term only involves powers of k . Don't forget you are talking to a computer when you use these online systems: of course Wolfram Alpha can use the Integral Test -- it can look up the improper integrals (or run a numerical calculation) internally...

    EDIT: I am withdrawing what I had written here earlier -- a trigonometric substitution will do the job. I used a tangent substitution instead and ended up integrating secant (with the upper limit changing to pi/2); your integral is analogous. Since the antiderivative involves logarithm of a function which goes to infinity as theta approaches pi/2 , the integral indeed diverges, so the absolute series fails the Integral Test. However, I found this by the Limit Comparison Test in a third of the time.

    You are hobbling yourself by avoiding them: they are equally powerful for testing absolute convergence to the Ratio Test; they work on series where the Ratio Test fails.

    You would use the A.S.T. after you've checked for absolute convergence. If the series is absolutely convergent, you don't need to bother. If it isn't, the A.S.T. is (relatively) easy to use next.

    There's a method for checking monotonic convergence if setting up an algebraic inequality is too difficult (or too much of a nuisance). Consider the absolute term [itex]b_{k} = \frac{k}{\sqrt{k^{4} + 2}}[/itex] and convert it to a function [itex]f(x) = \frac{x}{\sqrt{x^{4} + 2}} [/itex] . (We can see easily enough that [itex]\lim_{x \rightarrow \infty} f(x) = 0 [/itex].) Test to see if [itex] f'(x) < 0 [/itex] ; if the function decreases asymptotically to zero, then so does bk.
     
    Last edited: Sep 4, 2011
  6. Sep 5, 2011 #5
    You can try the comparison test by using k4 + 2 < 4k4 and then arrange it to look like the original series.
    Or if you really want to integrate it, start with the substitution u = x2
     
  7. Sep 5, 2011 #6
    Well my integral was wrong, I had 1/2 integral csc(theta) dtheta which gives me - 1/2*ln|csc(theta) + cot(theta)|, the correct antiderivative was suppose to be 1/2*ln|csc(theta)+cot(theta)|, without the negative sign I had included, or 1/2* ln|sqrt(k^4+2)/sqrt(2)+k^2/sqrt(2)|, with respect to k, which apparently is correct http://www.wolframalpha.com/input/?i=d%5B1%2F2*+ln|sqrt%28k^4%2B2%29%2Fsqrt%282%29%2Bk^2%2Fsqrt%282%29|%5D%2Fdk+%3D+k%2Fsqrt%28k^4%2B2%29

    1/2*integral sec(theta) dtheta = 1/2*ln|sec(theta)+tan(theta)| = 1/2*ln|sqrt(k^4+2)/k^2+sqrt(2)/k^2| seems to not be correct http://www.wolframalpha.com/input/?i=d%5B1%2F2*ln|sqrt%28k^4%2B2%29%2Fk^2%2Bsqrt%282%29%2Fk^2|%5D%2Fdk+%3D+k%2Fsqrt%28k^4%2B2%29
     
  8. Sep 5, 2011 #7
  9. Sep 5, 2011 #8

    dynamicsolo

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    Your integral is not incorrect. [EDIT: OK, found the sign error -- see post below] From the Pythagorean Identity,

    [tex]\sin^{2} \theta + \cos^{2} \theta = 1 \Rightarrow 1 + \cot^{2} \theta = \csc^{2} \theta \Rightarrow (\csc \theta + \cot \theta) (\csc \theta - \cot \theta) = 1 [/tex]
    [tex] (\csc \theta + \cot \theta) = \frac{1}{(\csc \theta - \cot \theta)} \Rightarrow \ln |\csc \theta + \cot \theta| = -\ln|\csc \theta - \cot \theta| . [/tex]

    Your trigonometric substitution [itex]\cot \theta = \frac{x^{2}}{\sqrt{2}} [/itex] leads to

    [tex]\int_{1}^{\infty} \frac{x dx}{\sqrt{x^{4}+2} } \longrightarrow -\frac{1}{2}\int_{\alpha}^{0} \csc \theta d\theta = -\frac{1}{2} \cdot \ln (\csc \theta - \cot \theta) \vert_{\alpha}^{0} = -\frac{1}{2} \cdot \ln (\csc \theta + \cot \theta) \vert_{0}^{\alpha} ,[/tex]

    where [itex]\alpha = \cot^{-1} \frac{1}{\sqrt{2}} [/itex] (we don't actually care what the exact value is, but only that it is some constant) and we have reversed the order of integration and changed the difference to a sum in the logarithmic term using the trigonometry identity. (And I know I should write a limit in the evaluation expression, instead of just an integration limit of "zero" -- indulge me: it's a shorthand...)

    [I might also remark that your integration will be a bit less agonizing to work out if you use [itex]\cot \theta = \frac{x^{2}}{\sqrt{2}} \Rightarrow -\csc^{2} \theta d\theta = \frac{2}{\sqrt{2}} x dx [/itex] .]

    Both the cosecant and cotangent function run off to positive infinity as theta approaches zero, so our integral diverges. When you transformed the evaluation back to the original variable, you had
    [itex]\lim_{a \rightarrow \infty} -\frac{1}{2} \cdot \ln (\frac{\sqrt{x^{4} + 2}}{\sqrt{2}} + \frac{x^{2}}{\sqrt{2}}) \vert_{1}^{a} [/itex] , where we again see that both terms in the logarithmic expression go off to infinity. (And you really want to have that sum in the logarithm, so you don't have to waste time thinking about an indeterminate difference of infinities...) [EDIT: found the sign error here -- see post below]

    So you were fine: you showed by the Integral Test that the series is not absolutely convergent.


    Note to Bohrok: I think that is a later copying error. GreenPrint has the correct result in the attachment back in post #1 [EDIT: oh well, there was a buried sign error...]


    I'm also going to make a comment about computer algebra/calculus systems (or looking up answers in the back of a textbook). This illustrates why it is important to know various identities and methods of simplification or transformation. I've had students come to me, occasionally on the verge of tears, because they'd spent a lot of time checking and re-checking their work and still weren't getting "the answer the book gives". There's typically about a half-percent chance (0.2 to 2%, depending on the lateness of the edition) that the "book answer" has a typo or is plain wrong; there is a much better chance that you did nothing wrong and that your answer is equivalent to the given one. That is something always worth checking: there can sometimes be multiple ways to write the same mathematical result.

    On computer systems, you have to deal with whatever way the designers/programmers decided to parse input expressions and "solve" for results. I've seen online integrators "stumped" by indefinite integrals that can be solved, simply because the code didn't "take apart" the integrand in a useful way. The Machine is not All -- it's a dumb robot that does calculus by "reading" expressions and looking up items in databases...
     
    Last edited: Sep 5, 2011
  10. Sep 5, 2011 #9
    I thought I was trying to evaluate the following integral

    lim a->infinity integral[1,a] k/sqrt(k^4+2) dk

    I didn't make a cot(theta)= x^2/sqrt(2) subsitution, from my triangle in the attachment I made a sine substitution (when I learned trig substitutions I taught myself by sort of just thinking about it and doing it myself without a book or a teacher so I sort of do it in a nontraditional sort of way as a result.)

    sin(theta) = opposite/hypotenuse = sqrt(2)/sqrt(k^4+2)

    lim a->infinity integral[1,a] k/sqrt(k^4+2) dk = lim a->infinity 1/sqrt(2)*integral[1,a] k*sin(theta)dk

    Also from the triangle I substituted for k
    cot(theta) = adjacent/opposite = k^2/sqrt(2), sqrt(2)*cot(theta) = k^2, k=+/- sqrt( sqrt(2)cot(theta) ) = +/- 2^(1/4)*cot(theta)^(1/2)

    The new integral

    lim a->infinity +/- 1/sqrt(2)*2^(1/4) integral[1,a] cot(theta)^(1/2)*sin(theta)dk

    Also from my triangle I substituted for dk to get the integral in terms of one variable theta,
    already established
    k=+/- 2^(1/4)*cot(theta)^(1/2)
    so taking the derivative with respect to theta
    dk/dtheta = +/- ( 2^(1/4) csc(theta)^2 )/( 2*cot(theta)^(1/2) )
    simplified
    dk/dtheta = +/- csc(theta)^2/( 2^(3/4) cot(theta)^(1/2) )
    so
    dk = +/- csc(theta)^2/( 2^(3/4) cot(theta)^(1/2) ) dtheta

    so the new integral
    lim a->infinity +/- 1/sqrt(2)*2^(1/4) *+/- 1/2^(3/4) integral[1,a] ( cot(theta)^(1/2)*sin(theta) csc(theta)^2 )/(cot(theta)^(1/2) ) dtheta

    simplified the constants and "canceled out cot(theta)^(1/2)/cot(theta)^(1/2)"
    lim a->infinity 1/2 integral[1,a] sin(theta) csc(theta)^2 dtheta

    simplified the trig functions
    lim a->infinity 1/2 integral[1,a] sin(theta)/sin(theta)^2 dtheta
    = lim a->infinity 1/2 integral[1,a] 1/sin(theta) dtheta
    = lim a->infinity 1/2 integral[1,a] csc(theta) dtheta

    which gave me...
    -1/2 lim a->infinity ln|csc(theta)+cot(theta)| evaluated from 1 to a
    = -1/2 lim a->infinity ln| sqrt(k^4+2)/sqrt(2) + k^2/sqrt(2) | evaluated from 1 to a

    Your saying that this is the correct antiderivative, -1/2 ln| sqrt(k^4+2)/sqrt(2) + k^2/sqrt(2) |, with the negative sign? For some reason I believe that it is to because I don't see were I went wrong, but wolfram alpha says the correct one is without the negative sign, 1/2 ln| sqrt(k^4+2)/sqrt(2) + k^2/sqrt(2) |, (I had a negative sign and wasn't suppose to, but I don't see what I'm doing wrong.)
     
    Last edited: Sep 5, 2011
  11. Sep 5, 2011 #10

    dynamicsolo

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    If you look at the fourth line of what is in your original written attachment, you'll see that we are using exactly the same substitution. You then made trouble for yourself by making an algebraic manipulation with square roots that led to a sign ambiguity; you then dropped the minus sign in the next line, which is not correct.

    Proceeding for your correct statement, [itex]\cot \theta = \frac{k^{2}}{\sqrt{2}} [/itex] , or [itex]k^{2} = \sqrt{2} \cot \theta [/itex] , you would then differentiate both sides to obtain [itex] 2k dk = -\sqrt{2} \csc^{2} \theta [/itex]. If you do it this way, there is no question about the presence of the minus sign. This makes your indefinite integral
    [tex]-\frac{1}{2} \int \csc \theta d\theta = -\frac{1}{2} [\ln | \csc \theta - \cot \theta |] .[/tex]

    We then apply the trig identity I described earlier to switch the logarithmic term to a sum (because we really don't want to deal with a difference in there when we do the evaluation), so our anti-derivative is [itex]\frac{1}{2} [\ln | \csc \theta + \cot \theta | ] .[/itex] If you want to transform this back to variable k , you will indeed get
    [tex]\frac{1}{2} \ln (\frac{\sqrt{k^{4}+2}}{\sqrt{2}} + \frac{k^{2}}{\sqrt{2}}) ,[/tex]

    just like Wolfram Alpha says. (And I just noticed I kept typing x2 inside the radical in my other post, so I'll go find that... FIXED)

    [after looking long over the last step] I see where our difference lies now. You omitted the minus sign in the derivative of cotangent by choosing the wrong square root where you consolidated your integral. The transformation I made changes the integration limits from x = 1 to [itex]x \rightarrow \infty[/itex] to the interval [itex]\theta = \alpha [/itex] to [itex]\theta \rightarrow 0[/itex] (alpha being the angle described earlier). So my result

    [tex]\lim_{t \rightarrow 0} -\frac{1}{2} \ln( \csc \theta - \cot \theta ) \vert_{\alpha}^{t} = \lim_{t \rightarrow 0} -\frac{1}{2} \ln( \csc \theta + \cot \theta ) \vert_{t}^{\alpha}[/tex]

    is correct. What happens in the evaluation? We have

    [tex][ -\frac{1}{2} \ln( \csc \alpha + \cot \alpha ) ] - [ \lim_{t \rightarrow 0} -\frac{1}{2} \ln( \csc t + \cot t ) ];[/tex]

    since alpha is in the first quadrant, the cosecant and cotangent are positive, the logarithm is positive, and the antiderivative value is a small negative number, while in the limit term, the cosecant and cotangent are running off to positive infinity, so the logarithm is as well, making the antiderivative value go off to negative infinity. But that term gets subtracted, so we get a divergent result for the integral, going to positive infinity, as it should (since the integrand function is above the x-axis).

    Where I erred was in the back-transformation to x (or k in your solution): I forgot that alpha "maps" back to 1 and zero back to "infinity", so I have to flip the sign on my antiderivative again, making it positive, as Wolfram says.

    This illustrates once again that sign errors are the #1 cause of math disasters (at all levels, including professionals)...
     
    Last edited: Sep 5, 2011
  12. Sep 5, 2011 #11
    So my mistake was in this step?
    k=+/- 2^(1/4)*cot(theta)^(1/2)
    dk/dtheta = +/- csc(theta)^2/( 2^(3/4) cot(theta)^(1/2) )

    because d[cot(x)]/dx=-csc(x)^2
    I thought that
    dk/dtheta = +/- (-1) csc(theta)^2/( 2^(3/4) cot(theta)^(1/2) )

    were the +/- was from when we solved for k and took a square root and got +/- and (-1) is from when the derivative of cotangent, I thought that +/-*-1 = +/- because (+1)(-1) = -1 and (-1)(-1) = +1, so I concluded that +/-(-1) was just +/-???

    I was suppose to just choose one of the square roots, I guess in this case the positive one, either the positive one or the negative one and not both? Why shouldn't I use both? Like is there any way to know which one to choose, I don't think I wouldn't have known to choose the positive square root.
     
  13. Sep 5, 2011 #12

    dynamicsolo

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    You can't just leave the +/- like that: since we are working with positive values of x , there is not going to be a sign ambiguity in the derivative of cotangent (there isn't for this function anyway, since it's the negative square of cosecant) or in the integral itself. It is generally advisable to avoid methods which leave unclear choices for the sign, or to have a way of testing the expression to determine which sign belongs there.

    For this integral, we know the general term of the series is positive for positive values of k , so the sum must be positive (and I should have been watching for this myself). Therefore, your improper integral must give, in this case, a positive divergent result. That is the final test that tells you which sign to choose here and that the sign choice must be negative.
     
    Last edited: Sep 5, 2011
  14. Sep 5, 2011 #13
    Hm... well that makes some more sense. Well thank you very much for your assistance =).
     
  15. Sep 5, 2011 #14

    dynamicsolo

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    You're quite welcome: I learn something when I have to troubleshoot my own mistakes as well...
     
  16. Sep 5, 2011 #15
    I always check the conditions for the integral test after evaluating the integral, for some reason, and strangely enough d[k/sqrt(k^4+2)]/dk = (sqrt(k^4+2)-(2k^4)/sqrt(k^4+2))/(k^4+2), is correct http://www.wolframalpha.com/input/?i=d%5Bk%2Fsqrt%28k^4%2B2%29%5D%2Fdk+%3D+%28sqrt%28k^4%2B2%29-%282k^4%29%2Fsqrt%28k^4%2B2%29%29%2F%28k^4%2B2%29, the problem however is that the derivative isn't negative until after about 1.1892071. For the integral test for a series from n=1 to infinity I thought the function that you take the derivative has to be decreasing from n=1 to infinity, this is not the case for this series. Is this a problem? Were you start the series doesn't determine if the series converges or diverges because it's the end values that do so, correct? So even though the function isn't decreasing on 1<= k < infinity I can still say that my conclusions based on the integral test are valid because the function is decreasing on say 2<= k < infinity?

    My book says
    Suppose f is a continuous, positive, decreasing function for x => 1 and let ak = f(k) for k = 1,2,3.... Then

    sigma[k=1, infinity] ak and integral[1,infinity] f(x)dx

    either both converge or both diverge. In the case of convergence, the value of the integral is not, in general, equal to the value of the series.

    So the function for the series is not decreasing for x => 1, so is our conclusions still valid???
     
  17. Sep 5, 2011 #16

    dynamicsolo

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    The requirement x ≥ 1 is a bit deceptive: what matters for convergence is what the "infinite tail" of the series does, not a finite interval from k = 1 to k = N . For example, in the Alternating Series Test, as long as the function is decreasing monotonically for k greater than some finite N (and the limit bk approaches zero, of course), the series will converge.

    There are a number of these convergence tests which read that way; and it is true that if the terms are decreasing for k ≥ N , then they are (eventually) decreasing for k ≥ 1 . You can have a series converge with the first few (or few thousand) terms actually increasing, as long as the terms become decreasing beyond some point and approach the zero asymptote "fast enough".
     
  18. Sep 5, 2011 #17
    Interesting, I thought so, Thank you very much.
     
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