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Property of sequence

  1. Oct 28, 2015 #1
    Is it possible to come up for a sequence property or sequence sum property for ΣnK=1K^-1

    If so, what other sequence properties that are not commonly seen are there?
     
  2. jcsd
  3. Oct 28, 2015 #2

    BvU

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    Well, it is known that $$\ln (1-x) = - \Sigma_{n=1}^\infty \ {x^n\over n} $$ for ##|x|<1## , so you have a diverging series here. I think I remember the progress of the sum is logarithmic, but I can't reproduce that offhand. Had to do with changing over from sum to integral for ##n## large enough...
     
  4. Oct 28, 2015 #3

    Samy_A

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    Not sure I understand the question, but might this help?
     
  5. Oct 28, 2015 #4

    Mark44

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    What do you mean by "sequence property" and "sequence sum property"? These are not terms that are commonly used with sequences or series.
     
  6. Oct 28, 2015 #5
    The link helps yes

    The question is simple... is there a formula (or possible to create one) for calculating k^-1 like there is for K^1, k^2 etc.

    cf4177ddff54c7cce35fce6ce061a167.png

    Edited:


    How to find the sum and not just the nth term hmm

    Not commonly used?

    Some basic commonly used terms in pre-calculus

    684c687cac8b7f8d644ba3a15a8291cb.png
     
    Last edited: Oct 28, 2015
  7. Oct 28, 2015 #6

    BvU

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    Nope, that is from 1 to n !
     
  8. Oct 28, 2015 #7

    Mark44

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    You're not "calculating k-1". The summation on the right side is the closed form of the sum in the middle.
    The integral above is an approximation of the sum of the finite series Hn. The integral does not give you the nth term. The nth term of the sum is just 1/n.
    I thought you meant specific properties of the finite sum you showed in post #1.
    The above are really properties of finite sums, not sequences (despite the title).
     
  9. Oct 28, 2015 #8
    I know this.. I'm using the image to [better] illustrate again the equation in question

    True

    Semantics aside it should be obvious considering the original post shows the Sigma notation. The actual question here still persists
     
  10. Oct 28, 2015 #9

    Mark44

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    Your question (quoted below) included a summation, but asked about properties of a sequence, which suggests that you aren't clear about the differences between a sequence and a sum. Your question was very unclear, so you can't just brush things off with "semantics aside".

    Here is what you wrote in post 1.
    If you have a sum such as ##a_1 + a_2 + \dots + a_n##, you can always write this in closed form as ##\sum_{i = 1}^n a_i##, but it might not be possible to write the associated infinite series as a function.
     
  11. Oct 28, 2015 #10
    Despite them having similar connotations and being shown explicitly it somehow remains vague?

    n is denoted above the sigma and not infinity implying that it is a finite series... considering the alternative "might not be possible" makes for a poor interpretation of what was meant.

    Could we please veer off the subject of semantics and focus on the arithmetic, we both clearly understand each-other, at least now. None of us are on some math standardizing committee that will get to decide any of this anyways nor is it of my own interest to discuss
     
  12. Oct 28, 2015 #11

    Mark44

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    A sequence and a sum (finite or infite) are two different things. I don't know what you mean by "similar connotations."
    Here is a sequence: ##\{1, 1/2, 1/4, 1/8, ...\}##
    Here is a (finite) series: ##\sum_{i = 0}^n \frac 1 {2^n}##
    The terms in a series come from a sequence, but the two concepts are very different.

    For certain kinds of series, such as geometric series (as in the example above), there are formulas for writing the sum in closed form. For an arbitrary series, there is no method that I'm aware of to allow rewriting the series in closed form.
    A question has meaning; i.e., semantic content. If the meaning of the question is not clear, it's difficult for anyone to answer the question.
     
    Last edited: Oct 28, 2015
  13. Oct 28, 2015 #12

    pwsnafu

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    Euler showed
    ##\sum_{k=1}^n \frac{1}{k} = \int_{0}^{1} \frac{1-x^n}{1-x} \, dx = \sum_{k=1}^n (-1)^k \frac{1}{k} {n \choose k} ##
    also we have
    ##H_n = \psi(n) + \gamma##
    where ##\psi(x) = \frac{\Gamma ' (x)}{\Gamma(x)}## is the digamma function and ##\gamma## is the Euler-Mascheroni constant.
     
  14. Oct 29, 2015 #13
    What is the derivative of

    2fbf92a60a1d639c61a9408acd536437.png

    Is it just ln(T(n)) ?

    so

    ##\sum_{k=1}^n \frac{1}{k} = ln((n-1)!)/((n-1)!) + \gamma## ??

    Yea plugging in the numbers doesn't work. Is that the correct derivative?
     
    Last edited: Oct 29, 2015
  15. Oct 30, 2015 #14

    Samy_A

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    No.

    I'm not sure what you mean by T(n), but as far as I know the best we have is what @pwsnafu already posted:

     
  16. Oct 30, 2015 #15

    Svein

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    The standard definition of the Euler-Mascheroni constant is: [itex]\gamma = \lim_{n\rightarrow \infty}( (\sum_{k=1}^{n}\frac{1}{k})-\ln(n)) [/itex]. Rearrange this, and you have [itex]\sum_{k=1}^{n}\frac{1}{k}\approx\ln(n)+\gamma [/itex].

    This expression can be manipulated in various ways, I shall just point out one result: In order for [itex] \sum_{k=1}^{n}\frac{1}{k}>100[/itex], you need [itex]\ln(n)\geq 100 [/itex] approximately. This means that [itex] n\geq e^{100}\approx 2.7\cdot 10^{43}[/itex]. So, while the sum diverges, it diverges very slowly.
     
  17. Oct 30, 2015 #16
    T(n) https://en.wikipedia.org/wiki/Gamma_function
    T'(n) https://en.wikipedia.org/wiki/Digamma_function https://en.wikipedia.org/wiki/Polygamma_function

    I believe I interpreted the first derivative T' wrong though in the above post

    I'm less interested in the infinite (divergent) series. Although I know in some parts of science they tend to try and set values for infinite (divergent) series if I remember correctly the sum of all natural numbers is understood as -1/16... but this isn't what interests me.

    I'm more interested in the finite series because I don't want to have to write out or use calculators to find the sum of something like
    [itex]\sum_{k=1}^{n}\frac{1}{k}[/itex].

    I tried plugging in the value for something like

    [itex]\sum_{k=1}^{4}\frac{1}{k}[/itex]. and the formula

    which is exactly 2.083333...33 or 25/12 (using expansion and exhaustion)

    [itex]\sum_{k=1}^{4}\frac{1}{k}\approx\ln(4)+\gamma [/itex].

    is about 1.9635

    So although this looks promising and useful (to an extent) it is not exact (within significant variance) and not exactly what I'm looking for but very interesting and a step in the right direction for sure.
     
    Last edited: Oct 30, 2015
  18. Oct 30, 2015 #17

    Samy_A

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    I may be missing something here, but I still don't see what T(n) should be.
    If you mean ##\Gamma (n)##, that is a little confusing.

    Finding such a formula would be a notable achievement, not sure it is possible.
     
  19. Oct 30, 2015 #18
    It's confusing to me as well, but it's used in the above function he posted:

    ##H_n = \psi(n) + \gamma##
    where ##\psi(x) = \frac{\Gamma ' (x)}{\Gamma(x)}## is the digamma function

    T and T' are used which the article about the digamma function refers to as this

    2fbf92a60a1d639c61a9408acd536437.png

    The gamma function

    But still confused on what that first derivative of it is ##\Gamma '(n)## .. since I don't actually know how to take derivatives of extremes.

    I think it would be too, both notable and potentially impossible.. at least with standard notation or even any notation. Unless someone has done it already, which would be nice to learn about as well.
     
  20. Oct 30, 2015 #19

    Samy_A

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    I really don't want to start about semantics again, but I still don't understand. Nowhere in the linked Wikipedia pages do I see a function T being used. Could you simply define this function to make things clear (for me, maybe I'm missing the obvious).

    The derivative of the ##\Gamma## function has been given by @pwsnafu in terms of the digamma function. No idea what "take the derivative of extremes" means.
    There is (as far as I know) no formula for ##\Gamma '(n)## using only "elementary" elements. See this thread for a integral formula for ##\Gamma'(z)##.
     
  21. Oct 30, 2015 #20
    For the love of all that is Euclidean I sincerely hope you're not talking about T(n) vs ##\Gamma (n)## considering one was just used over the other to avoid having to learn how to/use the forum API etc.

    It is at the very top of this page https://en.wikipedia.org/wiki/Gamma_function

    Derivative of extemems i.e. f(x)=(x-1)! what is f'(x)

    The conclusion in that thread "cannot be written in simpler ways"
    That's unfortunate, if it's entirely true -- at least for this application. Since that would mean it's basically useless at this point to use to find a summation formula for the finite series of

    [itex]\sum_{k=1}^{n}\frac{1}{k}[/itex].
     
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