Binomial series for $|z|=1$?

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The binomial series is given by
$$(*) quad (1-z)^-alpha=sum_k=0^inftyfracGamma(k+alpha)Gamma(alpha), k! , z^k; ,, |z|<1,$$
where $alpha in mathbb C$ is an arbitrary complex number. My question is that the identity $(*)$ is true for $|z|=1$ ?



In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$, i.e, $(1-xi)^-1=...; xi in mathbb S^1$.




calculus sequences-and-series complex-analysis binomial-coefficients




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  • 2




    The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
    – Lord Shark the Unknown
    Aug 2 at 10:32










  • if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
    – pointguard0
    Aug 2 at 10:33






  • 3




    For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
    – GEdgar
    Aug 2 at 10:36











  • In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
    – Z. Alfata
    Aug 2 at 10:40











  • @Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
    – Lord Shark the Unknown
    Aug 2 at 10:42














up vote
0
down vote

favorite
1












The binomial series is given by
$$(*) quad (1-z)^-alpha=sum_k=0^inftyfracGamma(k+alpha)Gamma(alpha), k! , z^k; ,, |z|<1,$$
where $alpha in mathbb C$ is an arbitrary complex number. My question is that the identity $(*)$ is true for $|z|=1$ ?



In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$, i.e, $(1-xi)^-1=...; xi in mathbb S^1$.




calculus sequences-and-series complex-analysis binomial-coefficients




share|cite|improve this question

















  • 2




    The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
    – Lord Shark the Unknown
    Aug 2 at 10:32










  • if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
    – pointguard0
    Aug 2 at 10:33






  • 3




    For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
    – GEdgar
    Aug 2 at 10:36











  • In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
    – Z. Alfata
    Aug 2 at 10:40











  • @Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
    – Lord Shark the Unknown
    Aug 2 at 10:42












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





The binomial series is given by
$$(*) quad (1-z)^-alpha=sum_k=0^inftyfracGamma(k+alpha)Gamma(alpha), k! , z^k; ,, |z|<1,$$
where $alpha in mathbb C$ is an arbitrary complex number. My question is that the identity $(*)$ is true for $|z|=1$ ?



In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$, i.e, $(1-xi)^-1=...; xi in mathbb S^1$.




calculus sequences-and-series complex-analysis binomial-coefficients




share|cite|improve this question













The binomial series is given by
$$(*) quad (1-z)^-alpha=sum_k=0^inftyfracGamma(k+alpha)Gamma(alpha), k! , z^k; ,, |z|<1,$$
where $alpha in mathbb C$ is an arbitrary complex number. My question is that the identity $(*)$ is true for $|z|=1$ ?



In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$, i.e, $(1-xi)^-1=...; xi in mathbb S^1$.





calculus sequences-and-series complex-analysis binomial-coefficients





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 10:43
























asked Aug 2 at 10:30









Z. Alfata

864413




864413







  • 2




    The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
    – Lord Shark the Unknown
    Aug 2 at 10:32










  • if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
    – pointguard0
    Aug 2 at 10:33






  • 3




    For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
    – GEdgar
    Aug 2 at 10:36











  • In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
    – Z. Alfata
    Aug 2 at 10:40











  • @Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
    – Lord Shark the Unknown
    Aug 2 at 10:42












  • 2




    The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
    – Lord Shark the Unknown
    Aug 2 at 10:32










  • if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
    – pointguard0
    Aug 2 at 10:33






  • 3




    For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
    – GEdgar
    Aug 2 at 10:36











  • In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
    – Z. Alfata
    Aug 2 at 10:40











  • @Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
    – Lord Shark the Unknown
    Aug 2 at 10:42







2




2




The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
– Lord Shark the Unknown
Aug 2 at 10:32




The convergence on the unit circle depends on what $alpha$ is, and what $z$ is.
– Lord Shark the Unknown
Aug 2 at 10:32












if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
– pointguard0
Aug 2 at 10:33




if $z in mathbfR$ and $|z| = 1$ then LHS is 0, while RHS is some positive sum.
– pointguard0
Aug 2 at 10:33




3




3




For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
– GEdgar
Aug 2 at 10:36





For example, the series for $(1-z)^-1$ diverges for $|z|=1$ but the series for $(1-z)^-1/2$ converges for $z=-1$.
– GEdgar
Aug 2 at 10:36













In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
– Z. Alfata
Aug 2 at 10:40





In my case, $zin mathbb S^1$ the unit circle in $mathbb C$ and $alpha in mathbb R$, precisely $alpha=1$.
– Z. Alfata
Aug 2 at 10:40













@Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
– Lord Shark the Unknown
Aug 2 at 10:42




@Z.Alfata When $alpha=-1$ then $Gamma(alpha)$ doesn't exist...
– Lord Shark the Unknown
Aug 2 at 10:42















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