Is there a power series with $0$ radius of convergence? 

Question: Is there a power series with $0$ radius of convergence?(doesn't converge anywhere)


I asked my teacher this question and he replied in negative, however he didn't mention any proof for this, saying it is obvious that the power series must converge somewhere.
I am not able to to grasp the "obvious" here. Is there any proof for this?
 A: Your teacher is wrong. Consider the series given by:
$$
\sum_{n=0}^\infty n! x^n
$$
The ratio test tells us that the series diverges: $\frac{n!x^n}{(n-1)!x^{n-1}} = n x$. As long as $|x| > 0$, there is some $N$ such that this is $ > 1$ for all $n > N$, so the series diverges.
A: A power series
$$
\sum_{n=1}^{\infty} a_n(x-a)^n
$$
is always going to converge for $x=a$.
In general the radius of convergence if the "size" of the interval where the series converges. A series will fall into one of three categories


*

*The series converges for all real numbers. We say here the radius os convergence is $\infty$

*The series converges on an interval from $a$ to $b$ (possibly including the endpoints). We say here that the radius of convergence is $b-a$. 

*The series converges only at one number $a$. We say here that the radius of convergence is $0$.


So there is always a radius of convergence. The set/interval where a series converges is always non-empty.
A: Yes, the radius of convergence can be $0$, e.g., $\sum n!x^n$. It will, of course, still converge at $x=0$, but nowhere else. 
A: Consider $\sum_{n=1}^\infty ṇ^nz^n$, $z\in \mathbb{C}$. The radius of convergence in this case is zero.
