[Largedisplaystyle sum_{n=1}^infty (-1)^n frac{H_n}{2n+1}=mathbf{G}-frac{pi}{2}ln(2)
]
(Largemathbf{Proof:})
(Largemathbf{Method~One:})
Using the relation (displaystyle H_{n} = int_{0}^{1} frac{1-x^n}{1-x} mathrm{d}x), we find that the series reduces to
[egin{align*} sum_{n=1}^infty (-1)^n frac{H_n}{2n+1} &= int_{0}^{1} frac{2}{1-x^2} left( frac{pi x}{4} - arctan x
ight) mathrm dx \ &= int_{0}^{1} frac{2}{1-x^2} left( arctan left( frac{1-x}{1+x}
ight) - frac{pi (1-x)}{4}
ight) mathrm dx \ &= int_{0}^{1} frac{2}{1-x^2} arctan left( frac{1-x}{1+x}
ight) mathrm dx - int_{0}^{1} frac{pi}{2(1+x)} mathrm dx end{align*}
]
For the former one, we use the substitution (displaystyle t = frac{1-x}{1+x}) to obtain that
[int_{0}^{1} frac{2}{1-x^2} arctan left( frac{1-x}{1+x}
ight) mathrm dx = int_{0}^{1} frac{arctan t}{t} mathrm dt = mathbf{G}
]
The latter one reduces to (displaystyle -frac{pi}{2} ln 2), so the conclusion follows.
[Largeoxed{displaystyle sum_{n=1}^infty (-1)^n frac{H_n}{2n+1}=color{Blue}{mathbf{G}-frac{pi}{2}ln(2)}}
]
(Largemathbf{Method~Two:})
[sum_{n=1}^infty (-1)^n H_n t^n = frac{-ln(1+t)}{1+t}
]
Let (t=x^2)
[sum_{n=1}^infty (-1)^n H_n x^{2n} = frac{-ln(1+x^2)}{1+x^2}
]
Integrating with respect to (x) and interchanging integral and summation we get
[egin{align*}
&sum_{n=1}^infty (-1)^n H_n int_0^1 x^{2n}mathrm dx = - int_0^1 frac{ln(1+x^2)}{1+x^2}mathrm dx\
&sum_{n=1}^infty (-1)^n frac{H_n}{2n+1}=-int_0^1 frac{ln(1+x^2)}{1+x^2}mathrm dx
end{align*}]
In the integral, substitute (displaystyle x= an( heta)):
[sum_{n=1}^infty (-1)^n frac{H_n}{2n+1} = 2int_0^{frac{pi}{4}}ln(cos heta) mathrm d heta
]
The remaining integral is evaluated using a fourier series:
[egin{align*}
sum_{n=1}^infty (-1)^n frac{H_n}{2n+1}
&= 2 int_0^{frac{pi}{4}}left( -ln(2)-sum_{j=1}^infty frac{(-1)^j cos(2j heta)}{j}
ight)mathrm d heta \ &=-frac{pi}{2}ln(2) -2sum_{j=1}^infty frac{(-1)^j}{j}int_0^{frac{pi}{4}}cos(2j heta) mathrm d heta \&= -frac{pi}{2}ln(2)+sum_{j=1}^infty frac{(-1)^{j+1}}{j^2}sin left( frac{pi j}{2}
ight) \
&=Largeoxed{displaystyle color{Blue}{mathbf{G}-frac{pi}{2}ln(2)}}
end{align*}]