Euler Sums系列（二）

\[\Large\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^2}=\frac{21}{16}\zeta(3)\]

\(\Large\mathbf{Proof:}\)
Let \(\displaystyle S_1=\sum_{n=1}^\infty \frac{H_n}{n^2}\) and \(\displaystyle S_2 = \sum_{n=1}^\infty(-1)^{n+1}\frac{H_n}{n^2}\). Then, our sum can be written as
\[\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^2} = \frac{S_1+S_2}{2}\]
We need to find \(S_1\) and \(S_2\).
1. Calculation of \(S_1\)
Note that
\[\frac{1}{k^2}= \int_0^1\int_0^1 (xy)^{k-1}\mathrm{d}x\mathrm{d}y~,~\frac{1}{n}= \int_0^1 z^{n-1} \mathrm{d}z\]
With the help of these, \(S_1\) can be calculated.
\[\begin{align*}
S_1 &= \sum_{k=1}^\infty \frac{1}{k^2}\sum_{n=1}^k \frac{1}{n}=\sum_{k=1}^\infty \sum_{n=1}^k \int_0^1\int_0^1 (xy)^{k-1}\mathrm dx \ \mathrm dy \int_0^1 z^{n-1} \mathrm dz \\ &= \sum_{n=1}^\infty \sum_{k=n}^\infty \int_0^1\int_0^1 (xy)^{k-1}\mathrm dx \ \mathrm dy \int_0^1 z^{n-1} \mathrm dz = \int_0^1 \int_0^1 \int_0^1 \left( \sum_{n=1}^\infty \frac{(xy z)^{n-1}}{1-xy}\right)\mathrm dx \ \mathrm dy \ \mathrm dz \\
&= \int_0^1 \int_0^1 \int_0^1 \frac{1}{(1-xy)(1-xyz)}\mathrm dx \ \mathrm dy \ \mathrm dz = \int_0^1 \int_0^1 \frac{\ln(1-xy)}{xy(xy-1)}\mathrm dx \ \mathrm dy \\
&=-\int_0^1 \int_0^1 \frac{\ln(1-xy)}{xy}\mathrm dx \mathrm dy -\int_0^1 \int_0^1 \frac{\ln(1-xy)}{1-xy}\mathrm dx \ \mathrm dy \\
&=-\int_0^1 \int_0^1 \frac{\ln(1-xy)}{xy}\mathrm dx \mathrm dy+\int_0^1 \frac{\ln^2(1-y)}{2y}\mathrm dy
\end{align*}\]
We have
\[\begin{align*}
\int_0^1 \int_0^1 \frac{\ln(1-xy)}{1-xy}\mathrm{d}x \ \mathrm{d}y &=- \sum_{n=1}^\infty \frac{1}{n} \int_0^1 \int_0^1 (xy)^{n-1}\mathrm{d}x \ \mathrm{d}y \\ &= -\sum_{n=1}^\infty \frac{1}{n^3}=-\zeta(3)
\end{align*}\]
and
\[\int_0^1 \frac{\ln^2(1-y)}{2y}\mathrm{d}y = \zeta(3)\]
So
\[S_1 = 2\zeta(3)\]
2. Calculation of \(S_2\)
This can be done in the same way as the previous one. We will use
\[\dfrac{(-1)^{k-1}}{k^2} = \int_0^1 (-x)^{k-1} \mathrm{d}x \int_0^1 z^{k-1} \mathrm{d}z = (-1)^{k-1} \int_0^1 \int_0^1 (xz)^{k-1} \mathrm{d}x \mathrm{d}z\]
Proceeding like the previous one we have
\[\begin{align*}S_2 &=\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac{1}{n} =\sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1\int_0^1 (-1)^{k-1} (xz)^{k-1}\mathrm{d}x\mathrm{d}z \int_0^1 y^{n-1} \mathrm dy\\& = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \dfrac{(-xyz)^{n-1}}{1+xz} \mathrm{d}x \mathrm{d}y \mathrm{d}z = \int_0^1 \int_0^1 \int_0^1 \dfrac1{(1+xz)(1+xyz)} \mathrm{d}x \mathrm{d}y \mathrm{d}z\\& = \int_0^1 \int_0^1 \dfrac{\ln(1+xz)}{xz(1+xz)} \mathrm{d}x \mathrm{d}z = \int_0^1 \int_0^1 \dfrac{\ln(1+xz)}{xz} \mathrm{d}x \mathrm{d}z - \int_0^1 \int_0^1 \dfrac{\ln(1+xz)}{1+xz} \mathrm{d}x \mathrm{d}z\\& = \int_0^1 \int_0^1 \dfrac{\ln(1+xz)}{xz} \mathrm{d}x \mathrm{d}z- \int_0^1 \dfrac{\ln^2(1+z)}{2z} \mathrm{d}z\end{align*}\]
Here
\[\begin{align*}
\int_0^1 \int_0^1 \dfrac{\ln(1+xz)}{xz} \mathrm{d}x \mathrm{d}z &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \int_0^1 \int_0^1 (x z)^{n-1} \mathrm{d}x \mathrm{d} z \\ &= \sum_{n=1}^\infty \frac{(-1)^n}{n^3}=\frac{3}{4}\zeta(3)
\end{align*}\]
and
\[\int_0^1 \frac{\ln^2(1+z)}{2z}\mathrm{d}z = \frac{\zeta(3)}{8}\]
So
\[S_2 = \frac{5}{8}\zeta(3)\]
3. Final answer
\[\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^2} = \frac{S_1+S_2}{2}= \frac{2\zeta(3)+\dfrac{5}{8}\zeta(3)}{2}=\Large\boxed{\color{blue}{\dfrac{21}{16}\zeta(3)}}\]
Note the identity
\[\psi(2n+2)+\gamma=H_{2n+1}\]
This gives us:
\[\sum_{n=0}^{\infty}\frac{\psi(2n+2)+\gamma}{(2n+1)^{2}}=\sum_{n=0}^{\infty}\frac{\psi(2n+2)}{(2n+1)^{2}}+\gamma\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2}}\]
The rightmost sum is rather famous, and evaluates to \(\displaystyle \frac{\gamma{\pi}^{2}}{8}\).
The left sum with the digamma term evaluates to
\[\Large\color{DarkGreen}{\sum_{n=0}^{\infty}\frac{\psi(2n+2)}{(2n+1)^{2}}=\frac{21}{16}\zeta(3)-\frac{\gamma{\pi}^{2}}{8}}\]

巴特西

Euler Sums系列（二）

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