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arXiv 0901.3452math.NTmath.CA
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Ramanujan求和与指数母函数$ \sum_{k=0}^{\infty}\frac{z^{k}}{k!}ζ^{\prime}(-k)$

Ramanujan Summation and the Exponential Generating Function $ \sum_{k=0}^{\infty}\frac{z^{k}}{k!}ζ^{\prime}(-k)$

B. Candelpergher, H. Gopalkrishna Gadiyar, R. Padma

摘要:

在第六章他的笔记本拉马努金介绍了一种方法,总结发散系列分配给该系列的价值相关联的欧拉-麦克劳林常数所产生的应用欧拉-麦克劳林求和公式的部分和系列。这种方法现在被称为拉马努金求和过程。本文计算了指数母函数$\sum_{n\geq 1}\log n e^{nz}$和$\sum_{n\geq 1}H_n^{(j)} e^{-nz}$的Ramanujan和,其中$H_n^{(j)}=\sum_{m=1}^n \frac{1}{m^j}$ .我们发现一个令人惊讶的关系,两个总和时,$j=1$从以下公式连接的衍生物的黎曼zeta函数在负整数的Ramanujan求和的分歧欧拉和$\sum_{n\ge 1} n^kH_n, k \ge 0$,其中$H_n= H_n^{(1)}$。此外,我们表示我们的结果在拉马努金求和的经典求和过程称为波莱尔和。

英文摘要:

In the sixth chapter of his notebooks Ramanujan introduced a method of summing divergent series which assigns to the series the value of the associated Euler-MacLaurin constant that arises by applying the Euler-MacLaurin summation formula to the partial sums of the series. This method is now called the Ramanujan summation process. In this paper we calculate the Ramanujan sum of the exponential generating functions $\sum_{n\geq 1}\log n e^{nz}$ and $\sum_{n\geq 1}H_n^{(j)} e^{-nz}$ where $H_n^{(j)}=\sum_{m=1}^n \frac{1}{m^j}$. We find a surprising relation between the two sums when $j=1$ from which follows a formula that connects the derivatives of the Riemann zeta - function at the negative integers to the Ramanujan summation of the divergent Euler sums $\sum_{n\ge 1} n^kH_n, k \ge 0$, where $H_n= H_n^{(1)}$. Further, we express our results on the Ramanujan summation in terms of the classical summation process called the Borel sum.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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