1. serie 1/n^5

Hello everyone!

How can I find the limit of $\displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}$. I know that it converges (easy to prove with the integral test), I also know that $\displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}= \zeta (5)$(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 $\displaystyle 10^-2$ so since $\displaystyle \zeta(5)=1.0369277551433699...$, I want to find 1.03)

Thanks

(I know the title of this thread is wrong but I can't change it know )

2. If you only need accuracy of $\displaystyle 10^{-2}$ why not just add the first few terms?

Also, if you want accuracy of $\displaystyle 10^{-2}$ you will actually want to find $\displaystyle 1.04$.

3. 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 $\displaystyle \sum_{n=1}^{\infty} 1/(4n^2+4n)=1/4$ because $\displaystyle \sum_{n=1}^{\infty} 1/(4n^2+4n)= \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n} -\frac{1}{n+1}=\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}- \dots -\frac{1}{n+1})=\frac{1}{4}$ ( \underbrace{}_{} doesn't work but $\displaystyle \frac{1}{n+1}=0$)
So is there something similar for $\displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}$ ?

4. 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 $\displaystyle \sum_{n=1}^{\infty} 1/(4n^2+4n)=1/4$ because $\displaystyle \sum_{n=1}^{\infty} 1/(4n^2+4n)= \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n} -\frac{1}{n+1}=\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}- \dots -\frac{1}{n+1})=\frac{1}{4}$ ( \underbrace{}_{} doesn't work but $\displaystyle \frac{1}{n+1}=0$)
So is there something similar for $\displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}$ ?
If you are trying to find the exact value of $\displaystyle \displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}$, then what you are trying to do is currently an open question in mathematics: summing reciprocals

5. Originally Posted by sunmalus
( \underbrace{}_{} doesn't work
$\displaystyle 1+2+\underbrace{3+4+5}+6+7$

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