Convergence question and degree of polynomial I'm currently teaching myself power series and Taylor's theorem for complex analysis and I'm having trouble answering questions of the following form:
$1)$ Suppose the power series $\sum_{k=0}^{\infty}a_kz^k$has radius of convergence 2. Call the value of the sum $f(z)$ and let $g(z)=\frac{f(z)}{1-z}$. 
$a)$ Explain how you know $g(z)$ has a representation of the form $g(z)=\sum_{n=0}^{\infty}b_nz^n$ valid at least for $|z|<1$.
$b)$ For each $n=0,1,2,3,$ find $b_n$ in terms of $a_0,a_1,a_2,a_3,...$
My approach for a:
$$g(z)=f(z)*\frac{1}{1-z}=f(z)*\sum_{n=0}^{\infty}z^n$$
 But since $f(z)= \sum_{k=0}^{\infty}a_kz^k$ 
Then: $$g(z)=\sum_{k=0}^{\infty}a_kz^k *\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}b_nz^n $$
This is from the fact derived by Taylor's theorem that $$\sum_{n=0}^{\infty}c_nz^n=\sum_{k=0}^{\infty}a_nz^n *\sum_{n=0}^{\infty}b_nz^n$$
Not sure how to do part b
$2)$ Suppose $f:\mathbb{C} \to \mathbb{C}$ is analytic on all $\mathbb{C}$, and $$|f^{(3)}(z)|<M$$ for all $z$ in $\mathbb{C}$. Show that $f$ is a polynomial and what can you say about the degree of $f$.
Is this problem also a result of Taylor's theorem? How do I approach it? Thank you.
 A: Part 1
I don't think your answer to this is correct. You need to justify why we can be sure that $g$ has a power series that converges if $|z|<1$, and you haven't done that.
You have correctly shown that $g$ is a product of two power series:
$$g(z)=\sum_{k=0}^{\infty}a_kz^k *\sum_{n=0}^{\infty}z^n$$
We also know that $f(z)=\sum_{k=0}^{\infty}a_kz^k$ has radius of convergence $R_1=2$, and $h(z) = \frac{1}{1-z}$ has radius of convergence $R_2=1$. Therefore, as a product of two power series, we know that the above expression for $g$ converges whenever $|z|< \min(R_1,R_2) = 1$. Note that we haven't yet shown that $g$ has a power series that converges when $|z|<1$ - that is something different.
However, power series converge uniformly within the radius of convergence, so, if $|z|<1$,we can multiply out the above expression to get:
$$\begin{align}
g(z) &= (a_0 + a_1z + a_2z^2 + \ldots)(1 + z + z^2 + z^3 + \ldots) \\
     &= a_0 + (a_1 + a_0)z + (a_2 + a_1+a_0)z^2 + \ldots
\end{align}$$ 
This shows that $g$ has a power series (that converges for $|z|<1$ by the above discussion), and also gives us an expression for the coefficients of the power series, i.e.:
$$b_n=\sum^n_{k=0}a_k$$ 
This simultaneously answers part 1b of your problem. (Note that the problem statement only asked for the case $n=0,1,2,3$.)
Part 2
As hinted at in the comments, you can apply Liouville's theorem to $f^{(3)}$ and integrate three times to deduce that $f$ is a polynomial of degree at most $3$.
