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I need to find the radius of convergence of the following series $$\displaystyle \sum_{n=0}^{\infty} \Big(\frac{z}{1+z}\Big)^n.$$

Here is my solution:

I know that $\displaystyle \sum_{n=0}^{\infty} w^n$ converges absolutely for $|w| < 1$, converges uniformly for $|w| \leq \delta < 1$ and diverges elsewhere. Applying this to the above problem, the given series converges absolutely iff $$\Bigg| \frac{z}{1+z} \Bigg| < 1 \iff \text{Re}(z) > \frac{-1}{2}$$

Hence, $\displaystyle \sum_{n=0}^{\infty} \Big(\frac{z}{1+z}\Big)^n$ converges absolutely for $|z|< \dfrac{1}{2}$ and uniformly on $\Big(|z| < \dfrac{1}{2}\Big)\setminus \Big\{\dfrac{-1}{2}\Big\}$

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  • $\begingroup$ How do you get uniform convergence on $(|z| <\frac 1 2)\setminus \{\frac {-1} 2\}$? $\endgroup$ – Kavi Rama Murthy Jan 18 '20 at 5:46
  • $\begingroup$ Sorry. It is not. Now, I feel it is absolutely convergent on $\Big(|z| \leq \frac{1}{2} \Big) \setminus \Big\{\frac{-1}{2}\Big\}$ and uniformly convergent on $\Big(|z| \leq \frac{1}{2} \Big) \setminus \Big\{\text{small arc near } z = \frac{-1}{2}\Big \} $, since it is compactly contained inside the set $\text{Re}(z) > \frac{-1}{2}$. $\endgroup$ – Ajay Kumar Nair Jan 18 '20 at 6:12
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Your main error is assuming that the area of convergence is a disk (with the border being 'unknown'), by insisting on a convergence radius!

That's only true for a power series, which has the form $\sum_{n=0}^{\infty}a_nz^n$, which your series is not. Other kinds of serieses have other shapes of their area of (absolute) convergence. As you correctly found out, your series has a half plane as area of absolute convergence:

$$\Re(z) > -\frac12$$

You then tried to interpret that result, saw the value $\frac12$ and tried to cram into your (wrong) notion that a convergance radius must exist. But that's not true, the area of absolute convergence is a half plane, not some disk!

Uniform convergence is another matter. Aside from solving

$$\left\vert\frac{z}{1+z}\right\vert \le \delta$$

for a given $0 < \delta < 1$, you need to find out if the map

$$z \to \frac{z}{1+z}$$

is uniformly continuous in that area as well.

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