Evaluation of series with binomial sums (Q2250859)
From MaRDI portal
scientific article
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Evaluation of series with binomial sums |
scientific article |
Statements
Evaluation of series with binomial sums (English)
0 references
21 July 2014
0 references
The author generalizes a summation formula from his not yet published manuscript ``Power series with binomial transforms and asymptotic expansions''. The main results are given in \S1 (Introduction and main results) in the form of three theorems: Theorem 1. Let \(F\) be the Laplace transform \[ F(s)=\int_0^{\infty}\,e^{-st}d\mu(t), \quad s\geq 0, \] of a finite measure \(\mu\) so that \(F'(0)\) is defined. Then for every complex number \(z\) with \(\text{Re}\,z>0\) we have \[ \sum_{n=1}^{\infty}\,{1\over n}\left\{\sum_{k=0}^n\,{n\choose k}(-1)^k F(zk)\right\}=-F'(0). \] Theorem 2. Under the assumptions of Theorem 1, we have \[ \sum_{n=1}^{\infty}\,{1\over n^2}\left\{\sum_{k=0}^n\,{n\choose k}(-1)^k F(zk)\right\}={\pi^2\over 6}F(0)-\sum_{n=1}^{\infty}\, {F(zn)\over n^2}+z\sum_{n=1}^{\infty}\,{F'(zn)\over n}. \] Theorem 3. Let \(F\) be the Fourier transform \[ F(s)=\int_{-\infty}^{\infty}\,e^{ist}d\mu(t) \] of a finite measure \(\mu\) such that \(F''(0)\) is defined. Then for every real number \(z\) we have \[ \sum_{n=1}^{\infty}\,{1\over n^2}\left\{\sum_{k=0}^n\, {n\choose k}(-1)^k [F(zk)+F(-zk)]\right\}=-{z^2\over 2}F''(0). \] This is followed by a section of examples (\S2, with 6 different types) covering quite diverse series. For instance \[ \sum_{n=1}^{\infty}\,{4^n\over n(2n+1)}\,{2n\choose n}^{-1}=2, \] \[ \sum_{n=1}^{\infty}\,{4^n\over n(2n+1)}\,{2n\choose n}^{-1} \left(1+{1\over 3}+\cdots+{1\over 2n+1}\right)=4. \] For \(0<x\leq 1\), \(a>0\) and \(r\) complex with \(\text{Re}\,r>0\) \[ \sum_{n=1}^{\infty}\,{1\over n}\left \{\sum_{k=0}^n\, {n\choose k}(-1)^k {x^k\over (k+a)^r}\right\}={r\over a^{r+1}}- {\log{r}\over a^r}, \] \[ \sum_{n=1}^{\infty}\, {1\over n^2}\left\{\sum_{k=0}^n {n\choose k}(-1)^k e^{-z^2k^2}\right\}={z^2\over 2}. \] The proofs are given in \S3 and the paper concludes with a list of 7 references.
0 references
summation of series
0 references
binomial coefficients
0 references
Fourier transform
0 references
Laplace transform
0 references
Hurwitz zeta function
0 references
dilogarithm
0 references
digamma function
0 references