特定的k=-3,求和的极限$n^k$

realnumber

k=-3,求极限$\sum_{i=1}^n i^k$.
又一般的k<-1,有和极限公式吗?
被一个小伙子问倒了
又k=-2也不晓得怎么推导.

kuing

`\zeta(3)` 大概是没初等表达式的。
`\zeta(2)=\pi^2/6` 我也不知道怎么推，搜了下有 zeta2（14种证明方法）

hbghlyj

kuing.cjhb.site/forum.php?mod=viewthread&tid=9638
Example 11.14. Let $g(z)=\cot (\pi z) / z^2$. By our discussion of the poles of $\cot (\pi z)$ above it follows that $g(z)$ has simple poles with residues $\frac{1}{\pi n^2}$ at each non-zero integer $n$ and residue $-\pi / 3$ at $z=0$.

Consider now the integral of $g(z)$ around the paths $\Gamma_N$ : By Lemma 11.13 we know $|g(z)| \leq C /|z|^2$ for $z \in \Gamma_N^*$, and for all $N \geq 1$. Thus by the estimation lemma we see that
$$
\left(\int_{\Gamma_N} g(z) d z\right) \leq C \cdot(4 N+2) /(N+1 / 2)^2 \rightarrow 0,
$$
as $N \rightarrow \infty$. But by the residue theorem we know that
$$
\int_{\Gamma_N} g(z) d z=-\pi / 3+\sum_{\substack{n \neq 0,-N \leq n \leq N}} \frac{1}{\pi n^2} .
$$
It therefore follows that
$$
\sum_{n=1}^{\infty} \frac{1}{n^2}=\pi^2 / 6
$$
Remark 11.15. Notice that the contours $\Gamma_N$ and the function $\cot (\pi z)$ clearly allows us to sum other infinite series in a similar way - for example if we wished to calculate the sum of the infinite series $\sum_{n \geq 1} \frac{1}{n^2+1}$ then we would consider the integrals of $g(z)=\cot (\pi z) /\left(1+z^2\right)$ over the contours $\Gamma_N$.
Remark 11.16. Note that taking $g(z)=$ $\left(1 / z^{2 k}\right) \cot (\pi z)$ for any positive integer $k$, the above strategy gives a method for computing $\sum_{n=1}^{\infty} 1 / n^{2 k}$ (check that you see why we need to take even powers of $n$ ). The analysis for the case $k=1$ goes through in general, we just need to compute more and more of the Laurent series of $\cot (\pi z)$ the larger we take $k$ to be.
One can show that $\zeta(s)=\sum_{n=1}^{\infty} 1 / n^s$ converges to a holomorphic function of $s$ for any $s \in \mathbb{C}$ with $\Re(s)>1$ (as usual, we define $n^s=\exp (s \cdot \log (n))$ where $\log$ is the ordinary real logarithm). As $s \rightarrow 1$ it can be checked that $\zeta(s) \rightarrow \infty$, however it can be shown that $\zeta(s)$ extends to a meromorphic function on all of $\mathbb{C} \backslash\{1\}$. The identity theorem shows that this extension is unique if it exists ${ }^{15}$. (This uniqueness is known as the principle of "analytic continuation".) The location of the zeros of the $\zeta$-function is the famous Riemann hypothesis: apart from the "trivial zeros" at negative even integers, they are conjectured to all lie on the line $\Re(z)=1 / 2$. Its values at special points however are also of interest: Euler was the first to calculate $\zeta(2 k)$ for positive integers $k$, but the values $\zeta(2 k+1)$ (for $k$ a positive integer) remain mysterious - it was only shown in 1978 by Roger Apéry that $\zeta(3)$ is irrational for example. Our analysis above is sufficient to determine $\zeta(2 k)$ once one succeeds in computing explicitly the Laurent series for $\cot (\pi z)$ or equivalently the Taylor series of $z \cot (\pi z)=i z+2 i z /\left(e^{2 i z}-1\right)$.

kuing

这帖还是移到高数区吧😌

tommywong

en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function

$\zeta(3)$叫做Apéry's constant
有條咁樣嘅公式似乎可以搵到$\zeta(2n+1)$嘅上界
$0=A_n\zeta(n)-B_n\pi^n+C_n S_-(n)+D_n S_+(n)$
$\displaystyle \zeta(3) = \frac{7}{180} \pi^3 - 2 \sum_{k=1}^\infty \frac{1}{k^3 (e^{2\pi k} - 1)}$
其它我就唔知喇

realnumber

谢谢各位,我转发给小伙看,了解下大概