I want to try and evaluate this interesting sum: ∑n=1∞1Γ(n+s) where 0≤s&lt;q

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I want to try and evaluate this interesting sum:  ∑n=1∞1Γ(n+s)  where 0≤s&amp;lt;q

Kingston Gates · Accepted Answer

We have∑n=1∞1Γ(n+s)=1Γ(s+1)(1+1s+1+1(s+1)(s+2)+…)One way to develop such a series is by successively integrating by parts the right hand side off(s)=∫01dζs(1−ζ)s−1F(ζ)where F(ζ) is some analytic function of ζ and ∫01 is shorthand for limϵ→0+∫01−ϵ. One findsf(s)=F(0)+F′(0)s+1+F″(0)(s+1)(s+2)+…For this problem we have F(n)(0)=1, so F(ζ)=eζ. Then∑n=1∞1Γ(n+s)=f(s)Γ(s+1)=1Γ(s+1)∫01dζs(1−ζ)s−1eζ=eΓ(s)∫01duus−1e−u=eΓ(s)γ(s,1)where γ(s,x) is the lower incomplete gamma function. Note that γ(s,x)=Γ(s)−Γ(s,x), where Γ(s,x) is the upper incomplete gamma function. Therefore,∑n=1∞1Γ(n+s)=e(1−Γ(s,1)Γ(s)as claimed.

terorimaox · Accepted Answer

I guess one starts by considering a more general sum:Eα,β(z)=∑n=0∞znΓ(αn+β)which is known as a Mittag-Leffler function. For the special case of α=1, the function satisfies a differential equation:zddzE1,s(z)+(s−1)E1,s(z)=∑n=0∞n+s−1Γ(n+s)=z∑n=0∞zn−1Γ(n−1+s)=1Γ(s−1)+zE1,sThis is an inhomogeneous equation of the first orderzy′(z)+(s−1−z)y(z)=1Γ(s−1)zddz(zs−1e−zy(z))=1Γ(s−1)zs−1e−zHencey(z)=1zs−1e−z(C−1Γ(s−1)∫z∞ts−2e−tdt)The integral on the right hand-side is known as incomplete Gamma function.Incidentally, the original series is also a hypergeometric series, meaning that E1,s(z) represents a hypergeometric function. Indeed:E1,s(z)=1Γ(s)F1(1;s;z)=1Γ(s)∑n=0∞(1)n(s)nzn=∑n=0∞znΓ(n+s)where (s)n=Γ(n+s)Γ(s) was used.

RizerMix · Accepted Answer

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I want to try and evaluate this interesting sum: \sum_{n=1}^\infty\frac{1}{\Gamma(n+s)} where 0\leq

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