- #1
Gregg
- 459
- 0
[tex] \sum _{n=1}^{\infty } \frac{1}{n^4} [/tex]
I can do 1/n^2 easily by using x^2 as a function but for this I try
[tex] x^4=\frac{1}{2\pi }\int_{-\pi }^{\pi } x^4 \, dx+\sum _{n=1}^{\infty } \left(\frac{2}{\pi }\int _0^{\pi }x^4\text{Cos}[n x]dx\right)\text{Cos}[n x] [/tex]
I arrive at:
[tex] \pi ^4=\frac{\pi ^4}{5}+\sum _{n=1}^{\infty } \frac{8 \left(-6+n^2 \pi ^2\right)}{n^4 } [/tex]
[tex] \frac{4\pi ^4}{5}=\sum _{n=1}^{\infty } \frac{8 \left(-6+n^2 \pi ^2\right)}{n^4 } [/tex]
f(x) = x^4 does not seem helpful to sum 1/n^4, maybe I need a polynomial in x of degree 4?
I can do 1/n^2 easily by using x^2 as a function but for this I try
[tex] x^4=\frac{1}{2\pi }\int_{-\pi }^{\pi } x^4 \, dx+\sum _{n=1}^{\infty } \left(\frac{2}{\pi }\int _0^{\pi }x^4\text{Cos}[n x]dx\right)\text{Cos}[n x] [/tex]
I arrive at:
[tex] \pi ^4=\frac{\pi ^4}{5}+\sum _{n=1}^{\infty } \frac{8 \left(-6+n^2 \pi ^2\right)}{n^4 } [/tex]
[tex] \frac{4\pi ^4}{5}=\sum _{n=1}^{\infty } \frac{8 \left(-6+n^2 \pi ^2\right)}{n^4 } [/tex]
f(x) = x^4 does not seem helpful to sum 1/n^4, maybe I need a polynomial in x of degree 4?