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Does anyone know an infinite series summation that is to 1/5 or 1/7?

  1. Dec 31, 2013 #1
    The title pretty much says it all, does anyone know infinite series summations that are equal to 1/5 or 1/7?
  2. jcsd
  3. Dec 31, 2013 #2


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    You can easily invent a geometric series that sums to any value you want.

    Or you could do something like the series expansion of ##\frac 1 5 \sin \pi##.
  4. Dec 31, 2013 #3
    Excellent! We can get those by changing the initial value and inputting different integers into the series.

    Do you know of any other ways to get to these values outside of geometric and trigonometric series? For example do we have a factorial series for calculating these values?
    Last edited: Dec 31, 2013
  5. Jan 1, 2014 #4


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    Any function whose value you can calculate and whose Taylor series can be expanded will give you a series. What is the purpose of this question?
  6. Jan 1, 2014 #5
    That won't get him 1/5, though. :tongue:

    I'm sure that was a typo. mesa, he probably meant ##\frac{\sin\frac{\pi}{2}}{5}##.
  7. Jan 1, 2014 #6


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    How about the trivial example where the first term in the series is 1/5 and all the remaining terms are zero? No need to get all fancy here guys!
  8. Jan 1, 2014 #7

    mesa, you should look back at your other thread on infinite series. Many of the same series (plural) apply here.
  9. Jan 1, 2014 #8


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    Get any sum that yields any finite number, multiply it by a "normalizing factor".

    [tex]e=\sum_0^\infty\frac 1 {n!}[/tex]


    [tex]\frac 1 5 = \sum_0^\infty\frac 1 {5en!}[/tex]
    Last edited: Jan 1, 2014
  10. Jan 1, 2014 #9
  11. Jan 1, 2014 #10
    To get an idea of what is out there. Do we have a 'general solution' outside of geometric series, trig, and normalizing factors for solving all whole integer fractions with '1' in the numerator through infinite series?

    Very good, it can be tough to spot errors late on New Years Eve... :)


    Yes they do, I was curious if we had more 'general' solutions for infinite series that would give whole integer fractions.

    That looks familiar :)

    Hmmm, that link doesn't seem to be working?
  12. Jan 1, 2014 #11


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    More of a brain-fart than a typo. I divided ##2\pi## by ##2## and got ##\pi##. Full marks for arithmetic, but zero marks for doing the wrong sum. :redface:
  13. Jan 1, 2014 #12


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    1/5 = 0.19999...

    1/7 = 0.142857142857...
  14. Jan 1, 2014 #13
    Nah. We all make mistakes or have lapses in genius from time to time. It's all good. :biggrin:

    What do you mean "more general"?
  15. Jan 1, 2014 #14
    I should probably have said, 'general solution' for an infinite series that can be used to get any whole integer fraction by changing the index of summation. For example, an infinite series that would be able to compute 1/5 and 1/7 simply by changing the index. Do you know if we have anything like this?
  16. Jan 2, 2014 #15
    If you want a series that converges to c, take a series that converges to b, and add this term: (c-b) to the front.
  17. Jan 3, 2014 #16
    That certainly would work much like Boreks suggestion of using a 'normalizing factor'.

    Do you know of any series that will equal all whole integer fractions by simply changing the lower limit of summation?
    Last edited: Jan 3, 2014
  18. Jan 3, 2014 #17
    Consider the sequence $$a_i=\{1,-1,\frac{1}{2},\frac{1}{2},-1,\frac{1}{3},\frac{1}{3},\frac{1}{3},-1,...\}.$$

    ##\displaystyle \sum_{n_0\leq i\leq n}a_i## should cover everything you want. :tongue:
  19. Jan 3, 2014 #18
    ...and I need to brush up on my writing comprehension skills, you guys are too clever :redface:
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