Okay, this is the editied version.

Another old exam question that I can't shake from my mind. I am searching for an intermediate form (most probably the Taylor series) such that it is possible to analytically continue the function $$f(z) = \sum_{n=0}^\infty \frac{z^n}{n+\frac{1}{2}}$$ outside of the unit circle. The next step will be to write the function $f\,$ in a hypergeometric form. Any thoughts or insight would be greatly appreciated.



Clearly the series converges in the unit disc $D(0;1)$; moreover it does converge on the boundary of the disc with the only exception of the point $1$ (for a known theorem due to E. Picard), hence the set of convergence is $\overline{D}(0;1)\setminus \{1\}$.

Now, you can manipulate the series as follows: $$\begin{split} \sum_{n=0}^\infty \frac{z^n}{n+\frac{1}{2}} &= 2\ \sum_{n=0}^\infty \frac{z^n}{2n+1}\\ &= \frac{2}{\sqrt{z}}\ \sum_{n=0}^\infty \frac{(\sqrt{z})^{2n+1}}{2n+1}\; ; \end{split}$$ remembering the Taylor expansion: $$\text{arctanh } w =\sum_{n=0}^\infty \frac{w^{2n+1}}{2n+1}$$ you conclude: $$\sum_{n=0}^\infty \frac{z^n}{n+\frac{1}{2}} = \frac{2}{\sqrt{z}}\ \text{arctanh } \sqrt{z}\; ,$$ or: $$\sum_{n=0}^\infty \frac{z^n}{n+\frac{1}{2}} =\frac{1}{\sqrt{z}}\ \ln \frac{1+\sqrt{z}}{1-\sqrt{z}}$$ which amounts to the same thing.

  • $\begingroup$ Cool, I can gain the hypergeometric form from this too. Thank you. $\endgroup$ – Henry Shearman Dec 8 '11 at 0:51

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