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Math Help - serie 1/n^5

  1. #1
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    serie 1/n^5

    Hello everyone!

    How can I find the limit of  \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}<br />
. I know that it converges (easy to prove with the integral test), I also know that  \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}= \zeta (5)<br />
(The Riemann zeta function) but I don't want to use this.
    Could you please tell me how to begin or the entire solution.
    (The accuracy must be of 10^-2 so since  \zeta(5)=1.0369277551433699..., I want to find 1.03)

    Thanks

    (I know the title of this thread is wrong but I can't change it know )
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  2. #2
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    If you only need accuracy of 10^{-2} why not just add the first few terms?

    Also, if you want accuracy of 10^{-2} you will actually want to find 1.04.
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  3. #3
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    Yep, but my question is rather how can I find 1.036... without adding terms and without using the zeta function.
    For instance it's easy to find  \sum_{n=1}^{\infty} 1/(4n^2+4n)=1/4 because  \sum_{n=1}^{\infty} 1/(4n^2+4n)= \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n} -\frac{1}{n+1}=\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}- \dots -\frac{1}{n+1})=\frac{1}{4} ( \underbrace{}_{} doesn't work but \frac{1}{n+1}=0)
    So is there something similar for \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}<br />
?
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  4. #4
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    Quote Originally Posted by sunmalus View Post
    Yep, but my question is rather how can I find 1.036... without adding terms and without using the zeta function.
    For instance it's easy to find  \sum_{n=1}^{\infty} 1/(4n^2+4n)=1/4 because  \sum_{n=1}^{\infty} 1/(4n^2+4n)= \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n} -\frac{1}{n+1}=\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}- \dots -\frac{1}{n+1})=\frac{1}{4} ( \underbrace{}_{} doesn't work but \frac{1}{n+1}=0)
    So is there something similar for \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}<br />
?
    If you are trying to find the exact value of \displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}, then what you are trying to do is currently an open question in mathematics: summing reciprocals
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  5. #5
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    Quote Originally Posted by sunmalus View Post
    ( \underbrace{}_{} doesn't work
    1+2+\underbrace{3+4+5}+6+7

    Double click to see LaTeX code
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