Find $a, b \in \mathbb{R}$ such that power series $ \sum_{n=1}^{\infty} \frac{\arctan n^a}{n^b} x^n$ converges. Find $a, b \in \mathbb{R}$ such that power series $$ \sum_{n=1}^{\infty} \frac{\arctan n^a}{n^b} x^n$$ converges.
I had problem in finding radius of convergence, more precisely I don't know how to find limit $$R = \limsup_{n\to \infty} \sqrt[n]{\frac{\arctan n^a}{n^b}}$$
I tried to write $\arctan n^a$ as $$\arctan n^a = \frac{\pi}{2} - \arctan{\frac{1}{n^a}}$$ and expand that into Maclaurain series, that is equal to $$\frac{\pi}{2} - (\frac{1}{n^a} - \frac{1}{3n^{3a}} + o(\frac{1}{3n^{3a}}))$$ From that I have got that $$R = \limsup_{n\to \infty} \sqrt[n]{\frac{\pi}{2n^b} - \frac{1}{n^{a+b}} + o(\frac{1}{n^{a+b}})}$$ but I am not sure what to do now. Any help will be welcome.
 A: Let's expand $\arctan$ as a power series around 0 as you already did.  As the power series for $\arctan$ has a radius of convergence of 1, we have to distinguish  cases:
$$
\arctan x = \begin{cases}
x + \Theta(x^3), & \text{ if } |x| < 1 \\
\pi/4, & \text{ if } x = 1 \\
\pi/2 + \Theta(x^{-1}), & \text{ if } x > 1 \\
\end{cases}$$
This means we have according cases $a$:
$$
\arctan(n^a) = \begin{cases}
n^a + \Theta(n^{3a}), & \text{ if } a < 0 \\
\pi/4, & \text{ if } a = 0 \\
\pi/2 + \Theta(n^{-a}), & \text{ if } a > 0 \\
\end{cases}$$
In the remainder, I will analyze the 1st case; the remaining cases are similar and you can work them out on your own. In the 1st case, we have:
$$\begin{align}
\frac{\arctan(n^a)}{n^b}x^n
&\stackrel{a<0}= \frac{n^a+\Theta(n^{3a})}{n^b}x^n \\
&= \big(n^{a-b} + \Theta(n^{3a-b}) \big)x^n \tag 1
\end{align}$$
Notice that the radius of convergence $R$ of
$$\sum_{k=1}^\infty n^c x^n \tag 2$$
is $R(c)=1$, regardless of the value of $c\in\Bbb R$.  Now the sum over $(1)$ consists basically of two such sums of kind $(2)$, each of which has radius of convergence of 1 regardless of $a$ and $b$. In particular, sum over $(1)$ converges if $|x|<1$ and diverges if $|x|>1$ regardless of $a$ and $b$.
The only case where $a$ and $b$ play a role are the cases $|x|=1$:

*

*If $x=1$, then sum of $(1)$ converges iff $a-b < -1$.

*If $x=-1$, then sum of $(1)$ converges iff $a-b \leqslant -1$.

As already said, the cases of $a\geqslant 1$ are similar.
A: It is simple using: ${\displaystyle r=\lim _{n\to \infty }\left|{\frac {c_{n}}{c_{n+1}}}\right|}$.
$$
{\frac {c_{n}}{c_{n+1}}} = \frac{\arctan n^a}{\arctan (n+1)^a} \cdot \frac{(n+1)^b}{n^b} = \frac{\arctan n^a}{\arctan (n+1)^a} \left(\frac{n+1}n\right)^b. 
$$
Now we calculate the radius of convergence: $r$:
If $b>0$,
$$
\lim_{n\to\infty} \left|{\frac {c_{n}}{c_{n+1}}}\right| = \frac{\lim_{n\to\infty}\arctan n^a}{\lim_{n\to\infty}\arctan (n+1)^a} \lim_{n\to\infty}\left(\frac{n+1}n\right)^a = \frac{\pi/2}{\pi/2} \left(\lim_{n\to\infty}\frac{n+1}n\right)^a=1^a=1. 
$$
If $b=0$,
$$
\lim_{n\to\infty}\left|{\frac {c_{n}}{c_{n+1}}}\right| = \frac{\arctan 1}{\arctan 1} \lim_{n\to\infty}\left(\frac{n+1}n\right)^a = \left(\lim_{n\to\infty}\frac{n+1}n\right)^a=1. 
$$
If $b<0$, the limit $\lim_{n\to\infty} \frac{\arctan n^a}{\arctan (n+1)^a}$ can be calculated by L'Hopital's rule and it would be $1$. Thus,
$$
{\lim_{n\to\infty}\left|{\frac {c_{n}}{c_{n+1}}}\right|} = 1. 
$$
Therefore, the radius of convergence is $1$ for any $a,b \in \mathbb R$.
