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Homework Help: Finding limits of summation

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data

    I have attached the question, along with the solution in the picture attached. This is one of the few questions I have encountered that I completely have no idea what the solution is trying to do...

    It's like they do not make any sense at all!

    Confused by
    1. The summation on the right - i have no idea what is the purpose of it

    2. The terms inside the brackets are also rather weird in terms of Mk-1 or Mk. The summation on the left - it starts from k = 2, 3, ..... which doesn't even match the terms in the brackets!

    2. Relevant equations

    3. The attempt at a solution

    I tried finding the d' Alembert's ratio but it gave me nonsensical answer of as long as x < 0 the series converges which is not true. If you compare the series to 1/n, the series diverges for x > -1

    Will appreciate it if anyone can help.
  2. jcsd
  3. Jul 9, 2012 #2


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    No attachment.
  4. Jul 10, 2012 #3
    ****, thanks for spotting my mistake man. Here's the attachment!

    Attached Files:

    Last edited: Jul 10, 2012
  5. Jul 11, 2012 #4
  6. Jul 11, 2012 #5


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    That solution is rather difficult to understand.

    I'll work on understanding it, then attempt to explain it.
  7. Jul 11, 2012 #6


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    Er if you've learned the ratio test, it would help you a lot here. Simply calculate :

    lim n->∞ |an+1/an|

    This will tell you whether your series is absolutely convergent, divergent or the test simply fails and you have to employ another test like the integral test or the alternating series test to find your answer. Also if you use this like so :

    lim n->∞ |an+1/an| < 1

    You can also determine for what values of x the series converges for as well as things like the radius of convergence.

    Hope this helps.
  8. Jul 20, 2012 #7
    bump...help anyone? I'm left with this last question for this chapter!

    p = lim ( n → ∞) [ (ln n+1)/(ln n) ]x < 1

    Then i tried to 'ln' both sides, which brings me:

    x * ln [ ln(n+1) ] < 0

    Which brings me a nonsensical answer of x < 0 which is not true, using a comparison test with 1/n .
  9. Jul 20, 2012 #8
    I'm guessing you aren't allowed to use the Cauchy Condensation Test, right? Here is a wikipedia link: http://en.wikipedia.org/wiki/Cauchy_condensation_test

    Basically, to me, it looks like the answer in the book is "proving" the condensation test for this one special case. I don't think you will get anywhere just doing a straight ratio test.
  10. Jul 23, 2012 #9
    It wasn't even written anywhere in the book! Thanks man, i'll have a go at the link.
  11. Jul 26, 2012 #10


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    I don't know if you're still interested, but I do understand the argument given in the link you gave.

    First, look at the values of Mk, for several value of k.

    k=0: M0 = 2, since ln(2) ≈ 0.693 and ln(1) = 0 .

    k=1: M1 = 3, since ln(3) ≈ 1.099 and ln(2) ≈ 0.693 .

    k=2: M2 = 8, since ln(8) ≈ 2.079 and ln(7) ≈ 1.946 .

    k=3: M3 = 21 .

    k=4: M4 = 55 .

    k=5: M4 = 149 .

    Looking at the nested summation:
    [itex]\displaystyle \large S_1=\sum_{k=1}^{\infty} \left(\sum_{r_k=1+M_{k-1}}^{M_k}\frac{1}{\left(\ln(M_k)\right)^X} \right)[/itex]​

    We see that when k = 1, r1 goes from 3, to 3:
    [itex]\displaystyle \frac{1}{\ln(M_1)}=\frac{1}{\ln(3)}<\frac{1}{\ln(n)}\,,[/itex] for n=2 .​

    When k = 2, r2 goes from 4, to 8:
    [itex]\displaystyle \frac{1}{\ln(M_2)}=\frac{1}{\ln(8)}<\frac{1}{\ln(n)}\,,[/itex] for n=3,...,7 .​

    When k = 3, r3 goes from 9, to 21:
    [itex]\displaystyle \frac{1}{\ln(M_3)}=\frac{1}{\ln(21)}<\frac{1}{\ln(n)}\,,[/itex] for n=8,...,20 .​


    There is an error in the last expression for S1. Mk - Mk-1 is only approximately equal to (1 - e-1)Mk, but they are equal in the limit k→∞ . So the final result is valid.

    There is a somewhat more straight forward way to achieve the desired result, using a modified version of this method.
  12. Jul 26, 2012 #11

    First of all, thank you for posting your solution. I have read everything and I understand every step, but i am still lost.

    I don't see how this answers the question and the purpose of this summation eludes me.. In short, I understand the steps, but I don't see how this is even related to the question at all

    The only thing i see is that this summation does the job of summating from k = 1 to k = ∞. But how did they even conceive of this in the first place?!
  13. Jul 26, 2012 #12


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    S1 < S.

    The ratio test for S1 shows that S1 diverges, for X considered here, i.e. X ≥ 1, which corresponds to x ≤ -1 .

    How did they come up with this? I don't know off hand, but it's basically regrouping the
    terms of the sum. A similar, but simpler, process can be used to show that [itex]\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}[/itex] diverges.
  14. Jul 26, 2012 #13

    Thank you, I now see how this solves to the question! This is one of the times when i look at the solution and get intrigued by how the person even came up with this..
  15. Jul 26, 2012 #14
    I figured this must be the thought process of the author:

    1. He first had the idea in mind to find a series S1 < S

    2. He saw that every term in the series gets smaller and smaller and sought to find a series that used the grouping method.

    2. Next he fiddled here and there and finally came up with the governing equation: ln(Mk) > k

    3. Then he tried evaluating certain values of k and tried to match the number of terms for each value of k, like for example behind ln 8 there's ln 3, ln 4, ..... ln 7. So the summation would be from Mk-1 + 1 to Mk.

    4. Then by ratio test, Voilà!

    What puzzles me still, is how he came up with ln Mk > k ..
  16. Jul 26, 2012 #15


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    It looks to me as if the person was relating the S sum to the sum [itex]\displaystyle \sum\frac{1}{n}\,,[/itex] so he based S1 on Ʃ(1/n).

    Looking at 1/ln(Mk) for several values of k appears to confirm this.
    k=1: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(3)}\approx\frac{1}{1.099}<\frac{1}{1}=\frac{1}{k}[/itex]

    k=2: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(8)}\approx\frac{1}{2.079}<\frac{1}{2}=\frac{1}{k}[/itex]

    k=3: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(21)}\approx\frac{1}{3.045}<\frac{1}{3}=\frac{1}{k}[/itex]

    k=4: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(55)}\approx\frac{1}{4.007}<\frac{1}{4}=\frac{1}{k}[/itex]

    k=5: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(149)}\approx\frac{1}{5.003}<\frac{1}{5}=\frac{1}{k}[/itex]

    k=6: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(404)}\approx\frac{1}{6.0014}<\frac{1}{6}=\frac{1}{k}[/itex]

    k=7: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(1097)}\approx\frac{1}{7.0003}<\frac{1}{7}=\frac{1}{k}[/itex]

    k=8: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(2981)}\approx\frac{1}{8.00001}<\frac{1}{8}=\frac{1}{k}[/itex]​

    As you can see, as k increases, ln(Mk) tends to get closer and closer to integer values.
  17. Jul 27, 2012 #16

    I see, thanks!! I will try to think harder next time :)
  18. Jul 27, 2012 #17


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    I doubt that this was a very easy solution to come up with.

    It certainly isn't all that easy to understand --- not very transparent.

    I'll try to get around to showing what I think is a somewhat clearer soliution in the next day or two.
  19. Jul 27, 2012 #18


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    The series, [itex]\displaystyle S=\sum_{n=2}^{\infty}\frac{1}{\left(\ln(n)\right)^R}\,,[/itex] can be thought of as a Left Riemann Sum for the improper integral [itex]\displaystyle \int_{2}^{\infty}\frac{1}{\left(\ln(x)\right)^R}\,dx\ .[/itex] Since the function, [itex]\displaystyle f(x)=\frac{1}{\left(\ln(x)\right)^R}[/itex] is monotonic decreasing, S is an upper bound for this integral.

    If we construct a Right Riemann Sum, S2 for this integral, then S2 is a lower bound & we have that [itex]\displaystyle S \ge\int_{2}^{\infty}f(x)\,dx\ge S_2\ .[/itex] If S2 does not converge for some value of R, then S does not converge for that same value of R.

    Construction of a Right Riemann Sum, S2:
    [itex]\displaystyle \text{Let }\ x_0=x_1=e\quad\to\quad f(x_1)=1/\left(\ln(e)\right)^R=1[/itex]

    In general, for k ≥ 1:

    [itex]\displaystyle \text{Let }\ x_k=e^k\quad\to\quad f(x_k)=1/\left(\ln(e^k)\right)^R=1/k^R\ \ \text{ and }\ \ \left(\Delta x\right)_k=e^k-e^{k-1}[/itex]

    This gives the following for S.

    [itex]\displaystyle S_2=\sum_{k=1}^{\infty}f(x_k) \left(\Delta x\right)_k=(e-2)+\sum_{k=2}^{\infty}\frac{e^k-e^{k-1}}{k^R}=(e-2)+\sum_{k=2}^{\infty}\frac{e^{k-1}(e-1)}{k^R}[/itex]​
    Applying the ratio test gives:
    [itex]\displaystyle \frac{\displaystyle \ \ \frac{e^{(k+1)-1}(e-1)}{(k+1)^R}\ \ }{\ \ \displaystyle \frac{e^{k-1}(e-1)}{k^R}\ \ }=e\left(\frac{k}{k+1}\right)^R\quad\to \ \ e\ \ \ \text{ as }\ k\to\infty[/itex]
    Thus S2 diverges and so does S.
  20. Jul 28, 2012 #19
    hmm that looks better than the previous solution. But why do you choose

    xk = ek , where Δx = ek - ek-1

    why can't you simply choose a simple one such that Δx = 1? Then it would resemble the previous S more..
  21. Jul 28, 2012 #20
    This may break the previous chain of thought, but I wanted to add a warning relating to SammyS's first proof using regrouping of terms. Generally, it is unsafe to use this technique. There exist examples of series where the order of summation may matter so that the limit of the series may not be well defined. There is a small section in Rudin that explains this phenomenon.
  22. Jul 28, 2012 #21


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    Hi "who". I think that only applies to series with both positive and negative terms which are not absolutely convergent. For a series like this with all positive terms I think that if it fails to converge for one ordering of the terms then it will fail for all.
  23. Jul 28, 2012 #22


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    Yes this is what I thought too Robert. The proof is so simple when using that theorem, that to be honest, I think it's probably easier to do it this way even if you had to prove the Cauchy Condensation Theorem first (for the [itex]2^n[/itex] case).
  24. Jul 28, 2012 #23
    Yes, in those scenarios it is safe. I was just putting it out there so that one might not get the notion that one can regroup terms in general. But absolute convergence is a must (in fact, that is the definition of absolute convergence).
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