Euler Sums系列（四）

\[\Large\displaystyle \sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1}=\mathbf{G}-\frac{\pi}{2}\ln(2)\]

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\(\Large\mathbf{Proof:}\)
\(\Large\mathbf{Method~One:}\)
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
\[\begin{align*} \sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1} &= \int_{0}^{1} \frac{2}{1-x^2} \left( \frac{\pi x}{4} - \arctan x \right) \mathrm dx \\ &= \int_{0}^{1} \frac{2}{1-x^2} \left( \arctan \left( \frac{1-x}{1+x} \right) - \frac{\pi (1-x)}{4} \right) \mathrm dx \\ &= \int_{0}^{1} \frac{2}{1-x^2} \arctan \left( \frac{1-x}{1+x} \right) \mathrm dx - \int_{0}^{1} \frac{\pi}{2(1+x)} \mathrm dx \end{align*}\]
For the former one, we use the substitution \(\displaystyle t = \frac{1-x}{1+x}\) to obtain that
\[\int_{0}^{1} \frac{2}{1-x^2} \arctan \left( \frac{1-x}{1+x} \right) \mathrm dx = \int_{0}^{1} \frac{\arctan t}{t} \mathrm dt = \mathbf{G}\]
The latter one reduces to \(\displaystyle -\frac{\pi}{2} \ln 2\), so the conclusion follows.
\[\Large\boxed{\displaystyle \sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1}=\color{Blue}{\mathbf{G}-\frac{\pi}{2}\ln(2)}}\]

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\(\Large\mathbf{Method~Two:}\)
\[\sum_{n=1}^\infty (-1)^n H_n t^n = \frac{-\ln(1+t)}{1+t}\]
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
\[\begin{align*} &\sum_{n=1}^\infty (-1)^n H_n \int_0^1 x^{2n}\mathrm dx = - \int_0^1 \frac{\ln(1+x^2)}{1+x^2}\mathrm dx\\ &\sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1}=-\int_0^1 \frac{\ln(1+x^2)}{1+x^2}\mathrm dx \end{align*}\]
In the integral, substitute \(\displaystyle x=\tan(\theta)\):
\[\sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1} = 2\int_0^{\frac{\pi}{4}}\ln(\cos \theta) \mathrm d\theta\]
The remaining integral is evaluated using a fourier series:
\[\begin{align*} \sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1} &= 2 \int_0^{\frac{\pi}{4}}\left( -\ln(2)-\sum_{j=1}^\infty \frac{(-1)^j \cos(2j\theta)}{j}\right)\mathrm d\theta \\ &=-\frac{\pi}{2}\ln(2) -2\sum_{j=1}^\infty \frac{(-1)^j}{j}\int_0^{\frac{\pi}{4}}\cos(2j \theta) \mathrm d\theta \\&= -\frac{\pi}{2}\ln(2)+\sum_{j=1}^\infty \frac{(-1)^{j+1}}{j^2}\sin \left( \frac{\pi j}{2}\right) \\ &=\Large\boxed{\displaystyle \color{Blue}{\mathbf{G}-\frac{\pi}{2}\ln(2)}} \end{align*}\]

转载于:https://www.cnblogs.com/Renascence-5/p/5458788.html