Find the radius of convergence of a series centered at a point a) Let $f(z) = \sum_{n≥0} z^n$. Find the radius of convergence of the series.
b) For all $t \epsilon \mathbb R$, $t&lt;1$, find the radius of convergence of the Taylor series of $f$ centered at $t$. (Hint: let $w = z-t$ and write $f(z) = f(w+t)$)

My attempts:
a) The radius of convergence is $\frac{1}{\alpha}$, where $\alpha := lim_{n→\inf}$ sup n$\sqrt{|a_n|}$.
But here $|a_n| = 1$ and so the limit (as defined above) is equal to $1$.
Hence $R=1$
b)$f(z) = f(w+t)$ = $\sum_{n≥0} {(t+w)}^n$ = $\sum_{n≥0} \sum_{k=0}^n \binom{n} {k} t^{n-k} w^k$
But $\binom{n} {k} = \frac{n!}{k!(n-k)!}$
Let $a_n = \frac{n!}{k!(n-k)!}$
And so lim as $n$ tends to infinity of $\frac{a_{n+1}}{a_n} = 1$.
And hence $R=1$
Are my attemps correct? Any other solutions please? Thank you
Your answer for (a) is correct. For (b) I think is correct and here is another way, let $1 &gt; t \in \mathbb{R}$.  Then the expansion of $f$ centered at $t$ is
$$ \sum_{n \geq 0} (z-t)^n $$
Which as you've already shown, converges for $|z-t|&lt;1$.  Let $z=x+iy$, then $f$ converges on the set
$$ \{z : |z-t|&lt;1 \} = \{ z : |z-t|^2&lt;1 \} = \{ x+iy : (x-t)^2 + y^2 &lt;1 \} $$
which is exactly the open disc of radius $1$ centered at $t$. So radius of convergence is $R=1$
Your answer for (a) is correct. For (b) I think is correct and here is another way, let $1 &gt; t \in \mathbb{R}$.  Then the expansion of $f$ centered at $t$ is
$$ \sum_{n \geq 0} (z-t)^n $$
Which as you've already shown, converges for $|z-t|&lt;1$.  Let $z=x+iy$, then $f$ converges on the set
$$ \{z : |z-t|&lt;1 \} = \{ z : |z-t|^2&lt;1 \} = \{ x+iy : (x-t)^2 + y^2 &lt;1 \} $$
which is exactly the open disc of radius $1$ centered at $t$. So radius of convergence is $R=1$