Prove $\frac{1^5}{1^5}+\frac{1^5-2^5}{1^5+2^5}+\frac{1^5-2^5+3^5}{1^5+2^5+3^5}+\frac{1^5-2^5+3^5-4^5}{1^5+2^5+3^5+4^5}+\cdots?$ On my previous page I proposed,
$$\frac{1^3}{1^3}+\frac{1^3-2^3}{1^3+2^3}+\frac{1^3-2^3+3^3}{1^3+2^3+3^3}+\frac{1^3-2^3+3^3-4^3}{1^3+2^3+3^3+4^3}+\cdots=3-\frac{\pi^2}{4}$$
Jack D'Aurizio proved the above identity in a simple manner.
He proposed the changing of cubes into the fifth power and he also offered the closed form of the identity.
$$\frac{1^5}{1^5}+\frac{1^5-2^5}{1^5+2^5}+\frac{1^5-2^5+3^5}{1^5+2^5+3^5}+\frac{1^5-2^5+3^5-4^5}{1^5+2^5+3^5+4^5}+\cdots=\\
15-\frac{\pi}{2}\left[3\pi+3\sec\left(\frac{\pi\sqrt3}{2}\right)-4\sqrt{3}\tan\left(\frac{\pi\sqrt3}{2}\right)\right]$$
Can anybody provide a proof for it?

We need some preliminary lemma.
$$ S_N=\sum_{n=1}^{N}n^5 = \frac{1}{12}N^2(N+1)^2 (2N^2+2N-1) \tag{1} $$
$$ P_N=\sum_{n=1}^{2N}(-1)^{n+1} n^5 = N^2(5-20 N^2-16 N^3) \tag{2}$$ 
$$ D_N=\sum_{n=1}^{2N-1}(-1)^{n+1} n^5 = N^2(5-20 N^2+16 N^3) \tag{3}$$ 
They give:
$$ S = \sum_{N\geq 1}\left(\frac{D_N}{S_{2N-1}}+\frac{P_N}{S_{2N}}\right)=\\\sum_{N\geq 1}\left(\frac{15-12 N^2 (5+4 N)}{(1-2 N)^2 \left(-1-4 N+8 N^2\right)}+\frac{15+12 N^2 (-5+4 N)}{(1+2 N)^2 \left(-1+4 N+8 N^2\right)}\right)\tag{4}$$
and such a series can be computed from partial fraction decomposition, the identity:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{5}$$
and the reflection formula for the $\psi=\frac{d}{dx}\log\Gamma$ function.
