7

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

math110
  • 93,304
  • 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. – Lucian May 05 '14 at 11:45
  • 2
    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...) – ajotatxe May 05 '14 at 11:56
  • @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 – math110 May 05 '14 at 12:01
  • Mathematica tells us that

    $$I=\sum_{n=1}^\infty\frac{1}{12n^4}\left(\pi^2-3(-1)^n\zeta\left(2,\frac{1+k}{2}\right)+3(-1)^n\zeta\left(2,\frac{2+k}{2}\right)\right).$$

    Here $\zeta(s,a)$ is the Hurwitz zeta function. Maybe this helps a bit?

     – 77474 May 05 '14 at 14:05
  • 1
    Some integral reps for $I$: $${\frac {1}{1080}},{\pi }^{6}+\int _{0}^{1}!{\frac {\ln \left( u \right) \mathrm{polylog} \left( 4,-u \right) }{1+u}}{du}={\frac {59}{30240}},{\pi }^{6}+\int _{0}^{\infty }!{\frac {t\mathrm{polylog} \left( 4,-{{\rm e}^{-t}} \right) }{1+{{\rm e}^{-t}}}}{dt}\approx1.063582242 $$ – Graham Hesketh May 10 '14 at 21:43
  • 5
    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 http://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. – George Lowther May 14 '14 at 00:01

0 Answers0