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Calculation of $$\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}$$

My attempt

\begin{align*} \sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2} &= -\sum_{n=1}^\infty\psi_1(n)\left(\frac{\log(2)}{2^n n}+\int_0^{\frac{1}{2}} x^{n-1}\log(x)dx\right) \\ &= \color{green}{-\log(2)\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}}\color{blue}{-\int_0^{1\over 2}\frac{\log(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)dx }\tag{1} \end{align*}

I don’t know how to evaluate $$\log(2)\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}$$ and $$-\int_0^{1\over 2}\frac{\log(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)dx$$

Martin.s
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1 Answers1

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Using Ali Shadhar's comment,

$$\color{blue}{\psi_1(n)=\zeta(2)-H^{(2)}_{n-1}}$$

$$S=\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}$$

$$S=\sum_{n=1}^{\infty}\frac{\zeta(2)-H^{(2)}_{n-1}}{2^nn^2}=\zeta(2)\sum_{n=1}^{\infty}\frac{1}{2^n n^2}-\sum_{n=1}^{\infty}\frac{H^{(2)}_{n-1}}{2^n n^2}$$

$$S=\zeta(2)S_1-S_2$$

Now, Using the definition of Dilogarithm,

$$\operatorname{Li}_2(z)=\sum_{n=1}^\infty\frac{z^n}{n^2} \underset{z=\frac12}\implies \operatorname{Li}_2\left(\frac12\right)=\sum_{n=1}^\infty\frac{1}{2^nn^2}$$

$$\boxed{S_1=\sum_{n=1}^{\infty}\frac{1}{2^n n^2}=\operatorname{Li}_2\left(\frac12\right)=\frac{1}{2}(\zeta(2)-\log^2(2))}$$

$$\boxed{S_2=\sum\limits_{n=1}^{\infty} \frac{H_{n-1}^{(2)}}{2^nn^2}=\frac1{16}\zeta(4)+\frac14\log(2)\zeta(3)-\frac14\log^2(2)\zeta(2)+\frac1{24}\log^4(2)}$$

$$S=\zeta(2)\left(\frac{1}{2}(\zeta(2)-\log^2(2))\right)-\left(\frac1{16}\zeta(4)+\frac14\log(2)\zeta(3)-\frac14\log^2(2)\zeta(2)+\frac1{24}\log^4(2)\right)$$

$$\boxed{\color{red}{S=\frac{1}{2}\zeta^2(2)-\frac{1}{4}\zeta(2)\log^2(2)-\frac1{16}\zeta(4)-\frac14\log(2)\zeta(3)-\frac1{24}\log^4(2)\approx 0.869761203}}$$

Which agrees with this

Luckily, there are answers that can explain the second summation better than I possibly can.

Similarly, one can also prove $\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}=\frac{1}{4}\zeta(3)+\frac{1}{6}\log^3(2)+\frac{1}{2}\zeta(2)\log(2)$

Amrut Ayan
  • 8,887