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Looking for comprehensive list (or link) of even power summations

  1. Jan 23, 2014 #1
    I have a function that results in 'exact' values for even powered summation series but it gives odd results for powers of '2' and '12+', how exciting! Unfortunately this also means the function is a far cry from a 'general solution'...

    Does anyone have a comprehensive list of power summations in the form of,

    $$\sum_{n=1}^{\infty} 1/n^m$$

    where 'm' is greater than the 10th power in exact form? (even of course)
  2. jcsd
  3. Jan 23, 2014 #2


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    That's just [itex] \zeta(m)[/itex]. The Bernoulli numbers can be calculated explicitly and can be written in terms of zeta values, see


    for the relationship between Bernoulli numbers and zeta values, and


    for a formula for calculating them. I also found this page


    which says it has a list of the Bernoulli numbers but you might want to check that it is correct.

    It's not clear what your post means when you say you have a function that results in exact values, but gives results that you think are wrong for almost all powers - in what manner is it supposed to be exact?
  4. Jan 23, 2014 #3
    The derived function will compute exact values of 'some' even powered summations. At the moment it will produce correct answers for,
    $$\sum_{n=1}^{\infty} \frac{1}{n^4}=\frac{Pi^4}{90}$$
    $$\sum_{n=1}^{\infty} \frac{1}{n^6}=\frac{Pi^6}{945}$$
    $$\sum_{n=1}^{\infty} \frac{1}{n^8}=\frac{Pi^8}{9450}$$
    $$\sum_{n=1}^{\infty} \frac{1}{n^{10}}=\frac{Pi^{10}}{93555}$$

    But for powers of '12' and higher the results don't look right and I know '2' is also incorrect (It is giving Pi^2/20, not Pi^2/6). I can easily fix the '2' but I need to see what is going on with the other end since it is best to work both sides of the problem.

    I could post the function although it is incomplete as far as being a 'general solution'. These discrepancies can likely be fixed but in order to do so I need a more comprehensive list of exact values for these power summations.

    On another note, thanks for the links!
  5. Jan 24, 2014 #4
    Hey Office_Shredder, I had a chance to run through these links although I am not clear on what this has to do with the question I posed, am I missing something or was my post just a little too hazy and that led you off track?
  6. Jan 24, 2014 #5


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    The first link gives the formula
    [tex] B_{2m} = (-1)^{m+1} \frac{2 (2m)!}{(2\pi)^{2m}} \left( \sum_{n=1}^{\infty} 1/n^{2m} \right)[/tex]
    which can be solved to give
    [tex] \sum_{n=1}^{\infty} 1/n^{2m} = (-1)^{m+1} B_{2m} \frac{ (2\pi)^{2m}}{2(2m)!} [/tex]

    So to calculate your sum on the left you need to evaluate the thing on the right, which is easy except for the B2m[/sum] part. I gave a link that will let you explicitly calculate it as a formula (if you wanted to compare your formula to the formula for the B's) and also a link which gives numerical evaluations of the first couple hundred.

    For example, from the last link B36 = -26315271553053477373/1919190, so letting m=18 above,
    [tex] \sum_{n=1}^{\infty} 1/n^{36} = 26315271553053477373/1919190 * \frac{ (2\pi)^{36}}{2(36)!} [/tex]

    We can confirm these are in fact the same number.

  7. Jan 24, 2014 #6

    Okay, it all makes sense now. I didn't notice in that first link to wikipedia that it was a summation for the far right term. Sometimes being so focused it is hard to spot even the obvious...

    I am pleased to see that,
    [tex] \sum_{n=1}^{\infty} 1/n^{12} = \frac{ 691Pi^{12}}{638512875} [/tex]

    I 'threw out' my solution as being anomalous because it had a 'non 1' numerated fraction for an answer (just like this does). I unfortunately left my notebook at home (probably crumpled up buried under a pile of sheets and pillows :P) although I will check this once I get home.

    So if this is in fact a 'general solution' then it is also a unique version of this Riemann Zeta function that doesn't require an input of your Bernoulli numbers, neat! (That is of course assuming the solutions for powers of 12+ are correct...)
  8. Jan 24, 2014 #7
    Office_Shredder, okay I checked it. The correct answer for n=12 is,

    $$\sum_{n=1}^{\infty} 1/n^{12} = \frac{ 691Pi^{12}}{638512875}$$

    My function gives,

    $$\sum_{n=1}^{\infty} 1/n^{12} = \frac{ 733Pi^{12}}{638512875}$$

    Close but not correct. I will work this from both ends and post when I have a solution. Thanks for the help.
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