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 $\Big(|z| < \dfrac{1}{2}\Big)\setminus \Big\{\dfrac{-1}{2}\Big\}$

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.