# serie 1/n^5

• Oct 2nd 2010, 09:32 AM
sunmalus
serie 1/n^5
Hello everyone!

How can I find the limit of $\sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}
$
. 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)
$
(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 :()
• Oct 2nd 2010, 09:40 AM
Prove It
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 $1.04$.
• Oct 2nd 2010, 10:24 AM
sunmalus
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 }}}
$
?
• Oct 2nd 2010, 02:56 PM
mr fantastic
Quote:

Originally Posted by sunmalus
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 }}}
$
?

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
• Oct 2nd 2010, 03:03 PM
undefined
Quote:

Originally Posted by sunmalus
( \underbrace{}_{} doesn't work :(

$1+2+\underbrace{3+4+5}+6+7$

Double click to see LaTeX code