Incomplete argument about the radius of convergence 
Suppose a sequence $(a_k)_{k \geq 0}$ such that $\sum_{k=0}^\infty
(-1)^k a_k$ converges and $\sum_{k=0}^\infty a_k$ diverges.
Determine the radius of convergence of the power series
$\sum_{k=0}^{\infty} a_k x^k$.

Let's define $f(x) = \sum_{k=0}^\infty a_k x^k$. As $f(-1) = \sum_{k=0}^\infty (-1)^k a_k$ converges and $f(1) = \sum_{k=0}^\infty a_k$ diverges, then if $R$ is the radius of convergence then $R \leq 1$ and diverges otherwise.
Someone told me it is incomplete as argument. What is the problem with this argument? What can I add to make it complete?
 A: 
Let's define $f(x) = \sum_{k=0}^\infty a_k x^k$. As $f(-1) = \sum_{k=0}^\infty (-1)^n a_k$ converges and $f(1) = \sum_{k=0}^\infty a_k$ diverges, then if $R$ is the radius of convergence then $R \leq 1$ and diverges otherwise.

I can't understand the clause that ends with "...and diverges otherwise" when it is not preceded by something that says that under some conditions something converges.
You wrote "then $R\le1$". That is consistent with $R=1$ and also with $R=0.4,$ etc., so it doesn't say $R=\text{some specified number.}$ So regardless of whether the argument is complete or incomplete in any sense, the conclusion is incomplete in that it does not fully answer the question.
To say that $\sum_{k=0}^\infty a_k(-1)^k$ converges is to say that $\sum_{k=0}^\infty a_k x^k$ converges when $x=-1.$ Since $-1$ is at a distance of $1$ from the center, the radius of convergence must be at least $1.$
If the radius of convergence were more than $1,$ then the power series would converge when $x=1,$ an you say it does not. Therefore the radius of convergence must be at most $1.$
Since it is at least $1$ and at most $1,$ it must be equal to $1.$
Thus we have the conclusion that $R=1,$ not just that $R\le 1.$
A: Since $\sum_{k=0}^{\infty}(-1)^n a_k$ converges, and a power series diverges everywhere outside the interval on convergence, we have $R\geq1$.  Since $\sum_{k=0}^{\infty}a_k$ diverges, and a power series converges everywhere inside the interval on convergence, we know $R\leq1$.  Therefore $R=1$.
I suppose this is what your friend meant.  Your argument is correct, but one can make a stronger statement.
