Radius of convergence I'm trying to understand how to handle power series who use floor or ceiling functions in their general term.
For example, the power series $\displaystyle{\sum_{k \geq 1} \left\lfloor \frac{2^k}{(k+1)^2}\right\rfloor}x^k$ is supposed to have a radius of convergence of $\frac{1}{2}$ but I fail to see how to handle this.
I see how the series converges for values less than $\frac{1}{2}$, but why does it diverge for larger values?
 A: You can also write $$a_k := \left \lfloor \frac{2^k}{(k+1)^2} \right \rfloor $$ and then it means that $$ a_k \leq \frac{2^k}{(k+1)^2} \leq a_k + 1 $$ and then multiplying by $|x|^k$ we get
$$ a_k |x|^k \leq \frac{2^k}{(k+1)^2} |x|^k \leq (a_k + 1) |x|^k $$
Now you can calculate the radius of convergence of the series $$\sum_{k = 1}^{\infty} \frac{2^k}{(k+1)^2} |x|^k $$ and it is equal to $1/2$.
And now you can conclude that the radius of convergence of the series $\sum a_k x^k$ is at least $1/2$ from the leftmost inequality. But using the rightmost inequality you can also see that the radius of convergence of the series $\sum a_k x^k$ cannot exceed $1/2$ because that would contradict the fact that the radius of convergence of the series $$\sum_{k = 1}^{\infty} \frac{2^k}{(k+1)^2} |x|^k $$ is $1/2$.
This is because if the series $\sum a_k x^k$ converges for some $x$ with $|x| > 1/2$ then the series $\sum (a_k + 1) x^k$ also converges because it is equal to the sum of the series $\sum a_k x^k + \sum x^k$, and then this would imply the convergence of the series $$\sum_{k = 1}^{\infty} \frac{2^k}{(k+1)^2} |x|^k $$
also for this $x$.
A: Notice that when $k$ is large enough, $\displaystyle{\left\lfloor \frac{2^k}{(k+1)^2}\right\rfloor \geq\frac{1}{2}\frac{2^k}{(k+1)^2}}$.  From this you can apply the root test to see that the radius of convergence is at most $\frac{1}{2}$.
