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tian27546西西 发表于 2013-11-9 04:22
我们想起定义harmonic sum identity
$$\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
其中$g^*(s)$ 是 $g(x).$ 的 Mellin transform
则我们有
$$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad
g(x) = \log\left(1-\frac{1}{3^x}\right).$$
Proving that $\frac{\pi}{4}$$=1-\frac{\eta(1)}{2}+\frac{\eta(2)}{4}-\frac{\eta(3)}{8}+\cdots$
Now recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1}\lambda_k f(\mu_k x); s\right)
= \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s}\right) f^*(s).$$
Put $$\lambda_k = (-1)^{k+1}, \quad \mu_k = k
\quad\text{and}\quad f(x) = \frac{1}{1+2x}$$
so that $$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \eta(s).$$
This yields
$$\mathfrak{M}\left(\sum_{k\ge 1} (-1)^{k+1} \frac{1}{1+2kx}; s \right)
= \frac{\eta(s)}{2^s} \frac{\pi}{\sin(\pi s)}.$$
The transform of $g(x)$ has fundamental strip $\langle 0,1\rangle$ and we may apply Mellin inversion to this transform to recover an expansion about infinity of the harmonic sum. This yields
$$\sum_{k\ge 1} (-1)^{k+1} \frac{1}{1+2kx} =
-\sum_{q\ge 1}
\operatorname{Res}\left(\frac{\eta(s)}{2^s} \frac{\pi}{\sin(\pi s)}/x^s; s=q\right)
= -\sum_{q\ge 1} \frac{\eta(q)}{2^q} \frac{(-1)^q}{x^q}.$$
It follows that
$$1 + \sum_{q\ge 1} \frac{\eta(q)}{2^q} \frac{(-1)^q}{x^q} =
1 + \sum_{k\ge 1} (-1)^k \frac{1}{1+2kx} =
\sum_{k\ge 0} (-1)^k \frac{1}{1+2kx}.$$
Finally put $x=1$ to obtain that
$$ 1 + \sum_{q\ge 1} (-1)^q\frac{\eta(q)}{2^q} =
\sum_{k\ge 0} (-1)^k \frac{1}{1+2k} = \frac{\pi}{4}.$$ |
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楼主: realnumber
hbghlyj
发表于 2023-5-9 01:16
