7
$\begingroup$

Question:

Find the sum $$I=\sum_{n=1}^{\infty}\dfrac{L_{n}(2)}{n^4}=?$$ where $$L_{n}(k)=1-\dfrac{1}{2^k}+\dfrac{1}{3^k}-\cdots+\dfrac{(-1)^{n-1}}{n^k}$$

since $$L_{n}(2)=1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots+\dfrac{(-1)^{n-1}}{n^2}$$ so $$I=\sum_{n=1}^{\infty}\dfrac{1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots+\dfrac{(-1)^{n-1}}{n^2}}{n^4}$$

Then I can't Continue

$\endgroup$
  • $\begingroup$ It's rather close to $1+\dfrac{\sqrt C}{10\cdot\ln(2+\sqrt3)}$ , where C is Catalan's constant, but I doubt that's the actual value. $\endgroup$ – Lucian May 5 '14 at 11:45
  • 2
    $\begingroup$ It seems a bit difficult... Can you specify which tools are supposed to be used? That is, what is the context of the problem? (Integration, number theory, Riemann's zeta function...) $\endgroup$ – ajotatxe May 5 '14 at 11:56
  • $\begingroup$ @ajotatxe,this problem is from china peopele,He very like research such this sum.and Now he can't find this.and He think it can find it the value $\endgroup$ – china math May 5 '14 at 12:01
  • 1
    $\begingroup$ Some integral reps for $I$: $${\frac {1}{1080}}\,{\pi }^{6}+\int _{0}^{1}\!{\frac {\ln \left( u \right) \mathrm{polylog} \left( 4,-u \right) }{1+u}}{du}={\frac {59}{30240}}\,{\pi }^{6}+\int _{0}^{\infty }\!{\frac {t\mathrm{polylog} \left( 4,-{{\rm e}^{-t}} \right) }{1+{{\rm e}^{-t}}}}{dt}\approx1.063582242 $$ $\endgroup$ – Graham Hesketh May 10 '14 at 21:43
  • 4
    $\begingroup$ You can rearrange it a bit as $$I=\sigma_a(2,4)+\frac{31}{32}\zeta(6)=\sigma_a(2,4)+\frac{31}{30240}\pi^6,$$ where $\sigma_a(2,4)$ is as in mathworld.wolfram.com/EulerSum.html. However, I haven't found any simple expression for $\sigma_a(2,4)$ in terms of standard constants, and suspect none is known. $\endgroup$ – George Lowther May 14 '14 at 0:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.