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

    Identifiers