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.