# Showing that a power series is convergent and finding radius of convergence

Let $$f(z) = \sum_{n≥0} a_n z^n$$ be a power series with radius of convergence $$R>0$$. Let $$t$$ $$\epsilon$$ $$D(0,R)$$.

a) Show that the series can be expressed in terms of the variable $$w=z-t$$ as $$\sum_{k≥0} b_k w^k$$ with $$b_k = \sum_{l≥k} a_l t^{l-k}$$ $$\binom{l} {k}$$

Moreover, show that the series defining $$b_k$$ is convergent.

b) Show that the radius of convergence of the series in (b) is at least $$R-|t|$$.

Can one show that the radius of convergence is exactly $$R-|t|$$?

My attempt for part a)

$$f(z) = \sum_{n≥0} a_n z^n = f(w+t) = \sum_{n≥0} a_n (t+w)^n = \sum_{k≥0} a_l\sum_{k=0}^l t^{l-k} w^k \binom{l} {k}$$ = $$\sum_{k≥0} b_k w^k$$ as desired.

$$b_k$$ is convergent because if $$c_l = a_l \frac{l!}{k!(l-k)!}$$, then lim as $$l$$ tends to infinity of $$\frac{c_{l+1}}{c_l}$$ = $$\frac{a_{l+1}}{a_l}$$. So the limit exists,and hence we have convergence.

Is my attempt for part a) correct? Any other suggestions please? Also, I tried solving part b) but couldn't get anywhere with it. Can someone show me how it is solved please? Thanks

Yes. The beginning of your attempt in part (a) is correct. Haven't verified your statement about the series for $$b_k$$ converging.

I would have said that since the original power series converges in a disk of radius $$R$$, you can find a disk of radius $$R-|t|$$ centered at $$t$$ and contained in the original disk. So, since the series for $$f(z)$$ converges in the larger disk, the series for $$f(w)$$, being the unique power series centered at $$w=0$$, must converge in the smaller disk.

• So by saying that, did you answer part a) or b) ?
– JOJO
Feb 19, 2020 at 21:43
• When I worked it out, got exactly what you got for the first part of (a). Didn't have time to look at the rest of your part (a), but since the series in $w$ centered at $w=0$ is unique, it must be a convergent series for $f(w)$ where analytic. I believe I also answered (b). The circle $|w|=R-|t|$ is the largest circle centered at $w=0$ and contained in the circle $z=R$. The series may or may not converge in a larger circle centered at $w=0$. There is at least one point $|z|\ge R$ where the series for $f(z)$ diverges. If there is such a point at $R \frac{t}{|t|}$, then this radius is max.
– mjw
Feb 19, 2020 at 23:17
• You've shown that $\frac{c_{l+1}}{c_l} ~ \frac{a_{l+1}}{a_l}$, but $b_k = \sum_{l\ge k} c_l$, so I don't think that says anything about the series converging.
– mjw
Feb 20, 2020 at 0:39
• then how do I show that $b_k$ is convergent? Please help
– JOJO
Feb 20, 2020 at 8:13
• Can we say anything about $b_k$ using the ratio test or root test?
– mjw
Feb 20, 2020 at 10:26