Find $a, b \in \mathbb{R}$ such that power series $ \sum_{n=1}^{\infty} \frac{\arctan n^a}{n^b} x^n$ converges.

Question

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.

Answer 1

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.

Answer 2

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 $a,b \in \mathbb R$.