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I wish to evaluate the sum
[tex]\sum_{k=n}^{\infty} \left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^k}{k+1}[/tex]
I think I can use Abel's theorem... just read: for |z|<1, we have
[tex]\frac{1}{1+z}\left( \frac{z}{1+z}\right) ^{n} = z^n\frac{1}{(1+z)^{n+1}} = (-1)^n\frac{z^n}{n!}\frac{d^n}{dz^n} \left( \frac{1}{1+z}\right) [/tex]
[tex]= (-1)^n\frac{z^n}{n!}\frac{d^n}{dz^n} \left( \sum_{k=0}^{\infty}(-1)^{k}z^{k} \right) = (-1)^n\frac{z^n}{n!}\sum_{k=n}^{\infty}(-1)^{k}\frac{k!}{(k-n)!}z^{k-n} = \sum_{k=n}^{\infty}(-1)^{n+k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k} [/tex]
here obtained is the identity
valid for |z|<1. Suppose that 0<x<1 and multiply the above identity by [tex](-1)^n[/tex] and then integrate over (0,x) to arrive at
to the latter sum apply Abel's Theorem to determine that
so that, by the previous identity, the value of the sum is given by
Okay, so www.integrals.com[/url] (also, [PLAIN]http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced#reply ) gives the integral in terms of the hypergeometric function 2F1, viz.
so here I need assistance, the sum above is to be again summed over, as in
where [tex] \zeta (s)[/tex] denotes Riemann's zeta function, and it should sum to [tex] -\log \sqrt{2\pi}[/tex]: and thoughts as to how to proceed from here?
[tex]\sum_{k=n}^{\infty} \left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^k}{k+1}[/tex]
I think I can use Abel's theorem... just read: for |z|<1, we have
[tex]\frac{1}{1+z}\left( \frac{z}{1+z}\right) ^{n} = z^n\frac{1}{(1+z)^{n+1}} = (-1)^n\frac{z^n}{n!}\frac{d^n}{dz^n} \left( \frac{1}{1+z}\right) [/tex]
[tex]= (-1)^n\frac{z^n}{n!}\frac{d^n}{dz^n} \left( \sum_{k=0}^{\infty}(-1)^{k}z^{k} \right) = (-1)^n\frac{z^n}{n!}\sum_{k=n}^{\infty}(-1)^{k}\frac{k!}{(k-n)!}z^{k-n} = \sum_{k=n}^{\infty}(-1)^{n+k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k} [/tex]
here obtained is the identity
[tex]\frac{1}{1+z}\left( \frac{z}{1+z}\right) ^{n} = \sum_{k=n}^{\infty}(-1)^{n+k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k} [/tex]
valid for |z|<1. Suppose that 0<x<1 and multiply the above identity by [tex](-1)^n[/tex] and then integrate over (0,x) to arrive at
[tex]\int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = \int_{z=0}^{x} \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k}dz = \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) \frac{x^{k+1}}{k+1} [/tex]
to the latter sum apply Abel's Theorem to determine that
[tex]\lim_{x\rightarrow 1^-} \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) \frac{x^{k+1}}{k+1} = \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1} [/tex]
so that, by the previous identity, the value of the sum is given by
[tex]\boxed{ \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1} = \lim_{x\rightarrow 1^-} \int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} }[/tex]
Okay, so www.integrals.com[/url] (also, [PLAIN]http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced#reply ) gives the integral in terms of the hypergeometric function 2F1, viz.
[tex]\lim_{x\rightarrow 1^-} \int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = \lim_{x\rightarrow 1^-} (-1)^k\frac{x^{k+1}}{k+1} _2F_1 (k+1,k+1;k+2;-x) = \frac{(-1)^k}{k+1} _2F_1 (k+1,k+1;k+2;-1) [/tex]
so here I need assistance, the sum above is to be again summed over, as in
[tex]\zeta ^{\prime} (0) =\sum_{n=0}^{\infty} (n+1)\left[ \log(n+1) -1\right] \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1} [/tex]
where [tex] \zeta (s)[/tex] denotes Riemann's zeta function, and it should sum to [tex] -\log \sqrt{2\pi}[/tex]: and thoughts as to how to proceed from here?
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