Find the sum of the series
\[\sum_{n=1}^{\infty}\frac{ 6 }{ n ^{6} }\]
do you realize 6/n^6 is n'th term of an GEOMETRIC series ?
what is a = 1st term = ..? what is common ratio = r = ..? then Sum of this series is just \(\Large S_n = \dfrac{a}{1-r}\)
to get a = 1st term, plug in n=1
but it says rounded of to the 3 decimal places
isn't a geometric supposed to be something like 6/(6^n) ?
yes, scratch that....it isn't geometric
this problem belongs to which topic ? any similar problems ?
I don't know the formula for the sum of 1/n^6, but here is the answer: https://www.wolframalpha.com/input/?i=sum+k%3D1+to+infinity+6%2Fk%5E6 :D
6.104 if you round to 3 decimal places
you guys dont know how to get that?
i might have came across such problems b4, but i nned to refresh what method was used....
I'm not at that level yet >.<
http://www.wolframalpha.com/input/?i=infinite%20sum%206%2Fn%5E6&t=crmtb01 Good luck finding this unless your name is Euler or Wolfram Alpha.
is this fourier series topic ?
Partially, although I think for the most part it's really just a clever use of power series. Here's a similar example: http://en.wikipedia.org/wiki/Basel_problem
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