[Largesum_{n=0}^infty frac{H_{2n+1}}{(2n+1)^2}=frac{21}{16}zeta(3)
]
(Largemathbf{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_0^1int_0^1 (xy)^{k-1}mathrm{d}xmathrm{d}y~,~frac{1}{n}= int_0^1 z^{n-1} mathrm{d}z
]
With the help of these, (S_1) can be calculated.
[egin{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_0^1int_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_0^1int_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}
ight)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
[egin{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 = 2zeta(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
[egin{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_0^1int_0^1 (-1)^{k-1} (xz)^{k-1}mathrm{d}xmathrm{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}zend{align*}
]
Here
[egin{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{2zeta(3)+dfrac{5}{8}zeta(3)}{2}=Largeoxed{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}}+gammasum_{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
[Largecolor{DarkGreen}{sum_{n=0}^{infty}frac{psi(2n+2)}{(2n+1)^{2}}=frac{21}{16}zeta(3)-frac{gamma{pi}^{2}}{8}}
]