2

show that $$\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln{n}}{n^2}+\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln^2{n}}{n}=\dfrac{\pi^2}{12}\ln{2}+\ln^2{2}\left[\gamma-\dfrac{1}{3}\ln{2}\right]?$$

where the $\gamma$ is Eluer constant.

I remember this sum is old: $$\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln{n}}{n^2}=\dfrac{\pi^2}{12}\ln{2}?$$ and think the general case $$\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln^p{n}}{n^q}=?$$

I have found this :link use idea: since $$f(x)=\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^x}$$ then $$f'(x)=-\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln{n}}{n^x}$$ $$f''(x)=\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln^2{n}}{n^x}$$ $$f'''(x)=\sum_{n=1}^{\infty}\dfrac{(-1)^n\ln^3{n}}{n^x}$$ $$\cdots$$ But Now How find this value $f'(1),f'(2),f''(1),f''(2),\cdots$?

maybe have some paper research it? Thank you

Community
  • 1
math110
  • 93,304
  • 1
    $$\zeta(x)=\sum_1^\infty\frac1{n^x}\qquad,\qquad \eta(x)=\sum_1^\infty\frac{(-1)^{n+1}}{n^x}=\Big(1-2^{1-x}\Big)\cdot\zeta(x)$$ $$=>\qquad\sum_1^\infty(-1)^{n+p+1}\frac{\ln^pn}{n^x}=\eta^{(p)}(x)= \frac{d^p}{dx^p}\bigg[\Big(1-2^{1-x}\Big)\cdot\zeta(x)\bigg]$$ See Riemann $\zeta$ function and Dirichlet $\eta$ function for more details. – Lucian Apr 07 '14 at 12:22

0 Answers0