$∑(-1)^k/(3k+1)$

hbghlyj

\begin{align*}\frac2π\int_0^π\left(\frac\pi4\operatorname{sgn}(x)\right)\sin(n x)\rmd x&={1 -\cos(π n)\over2n}\\&=\begin{cases}0&n\equiv0\pmod2\\\frac1n&n\equiv1\pmod2\end{cases}\end{align*}因此$\frac\pi4\operatorname{sgn}(x)$的Fourier sine series为 $$ f(x) = \sum_{k=0}^{\infty}\frac{\sin((1+2k)x)}{1+2k} $$在$x=\pi/2$处计算得$\sum_{k=0}^{\infty} (-1)^k/(1+2k)= \pi/4$ u[x_]:=Sum[Sin[(1+2k)x]/(1+2k),{k,0,∞}]; Plot[u[x], {x, 0, 2 Pi}] u[Pi/2] Copy the Code

为了证明：
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{3k+1} =\frac{\sqrt{3} \pi + \log(8)}9$$
仿照上面，构造$f(x)$$$ f(x) = \sum_{k=0}^{\infty}\frac{\sin((1+2k)x)}{1+3k}$$ 要计算$f(x)$在$\pi/2$的值。我到这卡住了，因为我不知道$f(x)$是哪个函数的Fourier sine series

hbghlyj

u[x_]:=FullSimplify[Sum[Sin[(1+2k)x]/(1+3k),{k,0,∞}]]; Plot[u[x], {x, 0, 2 Pi}] u[Pi/2]

$f(x)$用Mathematica的FullSimplify算得

1. 1/2 I E^(-I x) (Hypergeometric2F1[1/3, 1, 4/3, E^(-2 I x)] - E^(2 I x) Hypergeometric2F1[1/3, 1, 4/3, E^(2 I x)])

Copy the Code

好复杂

hbghlyj

这道题好像用Fourier series很难办，但是可以逐项积分：
https://math.stackexchange.com/questions/1482529
https://math.stackexchange.com/questions/960944
下面是AOPS的类似题：$$\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}=\frac{1}{3}-\frac{1}{7}+\frac{1}{11}-\frac{1}{15}+\cdots $$
We have,
$$x^2-x^6+x^{10}+\dots=\frac{x^2}{1+x^4}\tag1$$
Now, the radius of convergence of this power series is $1$.
So, integrating (1) from $0$ to $x\;(|x|<1)$ we get,
$$\frac{x^3}{3}-\frac{x^7}{7}+\dots=\int_{0}^{x}\frac{t^2}{1+t^4} dt$$ whose radius of convergence is $1$ and also at $x=1$ series converges by AST.
So, $\sum_{k=0}^{\infty}\frac{(-1)^k}{4k+3}=\lim_{x\to 1^{-}}\int_{0}^{x}\frac{t^2}{1+t^4} dt=\frac{\pi+2\ln(\sqrt{2}-1)}{4\sqrt{2}}$

tommywong

$\displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{ak+b}x^{ak}
=\frac{1}{x^b}\int_0^{x} \sum_{k=0}^\infty (-1)^k t^{ak+b-1} dt
=\frac{1}{x^b}\int_0^{x} \frac{t^{b-1}}{1+t^a} dt$

wolframalpha.com/input?i=int 1/(1+x^3)

$\displaystyle \int_0^x \frac{1}{1+t^3} dt= \frac{\ln\dfrac{x^2+2x+1}{x^2-x+1}+2\sqrt{3}\arctan\dfrac{2x-1}{\sqrt{3}}-2\sqrt{3}\arctan\dfrac{-1}{\sqrt{3}}}{6}$

$\displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{3k+1}
=\frac{\ln4+2\sqrt{3}\arctan\dfrac{1}{\sqrt{3}}-2\sqrt{3}\arctan\dfrac{-1}{\sqrt{3}}}{6}
=\frac{\ln8+\sqrt{3}\pi}{9}$