0
$\begingroup$

Question

Let $g(z):=\sum_{n=0}^\infty n(z+1)^n$.
Where does $g$ converge, what is its sum and where is it analytic?

My attempt

Let $\rho$ be its radius of convergence and so $\rho = \lim_{n\rightarrow\infty} \left|\frac{a_n}{a_{n+1}}\right| = \lim_{n\rightarrow \infty}\left|\frac{n}{n+1}\right| = 1.$
Hence, the radius of convergence is $1$, with centre $z_0 = -1$, so $g$ converges when $z$ is in the open-ball $B(-1,1)$.

Now I'm unsure how to answer the analytic function. Surely it is analytic in the ball, but how about on the boundary and outside the ball?

This is my attempt at finding the sum.
$$g(z) = 0 + (z+1)^1 + 2(z+1)^2 + 3(z+1)^3 + \ldots \\ = 2(z+1) + 3(z+1)^2 + 4(z+1)^3 + \ldots - ((z+1)+(z+1)^2 + \ldots) \\ = \sum_{n=1}^\infty (n+1)(z+1)^n - \sum_{n=1}^\infty (z+1)^n$$
Noting that both sums converge inside this ball mentioned previously, so it is valid (I think?) to split the sum into these two parts.

The first sum is equal to
$$f(z):= \sum_{n=1}^\infty (n+1)(z+1)^n = \frac{\mathrm{d}}{\mathrm{d}z}\left(\sum_{n=1}^\infty (z+1)^{n+1}\right)\\ = \frac{\mathrm{d}}{\mathrm{d}z}\left(\frac{(z+1)^2}{1-(z+1)}\right) = \frac{\mathrm{d}}{\mathrm{d}z}\frac{z^2 + 2z + 1}{-z} \\ = \frac{\mathrm{d}}{\mathrm{d}z} \left(-z - 2 - \frac{1}{z}\right) = -1 + \frac{1}{z^2}$$ and the second sum equals $-1 -\frac{1}{z}$.
So $g(z) = -1 + \frac{1}{z^2} + 1 + \frac{1}{z} = \frac{1}{z} + \frac{1}{z^2}$

$\endgroup$
  • 2
    $\begingroup$ For a function to be analytic at a point, it must be defined in a neighbourhood of that point. Does your open disc $D(-1,1)$ contain a neighbourhood of any point of its boundary or exterior? $\endgroup$ – Lord Shark the Unknown Aug 16 '17 at 10:22
  • $\begingroup$ No because $D(-1,1)$ is an open ball by defn (choice because I wasn't sure if it converges on its boundary), so it wouldn't contain its boundary or exterior $\endgroup$ – Twenty-six colours Aug 16 '17 at 10:40
  • $\begingroup$ $B(-1,1)$ or $B(-2,0)$ ? $\endgroup$ – Nosrati Aug 16 '17 at 10:43

Your Answer

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

Browse other questions tagged or ask your own question.