Ytukyg

I need to study the convergence of the series $sum_{n=1}^{infty}arctan(frac{2x}{x^2+n^4})$.

I thought about 2 ways it can be done.

One way is to try and calculate the partial sum of the series so we can see what is its limit.

Another way is to use the fact that if we have $sum_{n=1}^{infty}f_n(x)$ and we find a sequence $a_n$ so that

$|f_n(x)| leq a_n$

and

$sum_{n=1}^{infty}a_n$ is convergent

then $sum_{n=1}^{infty}f_n(x)$ is convergent.

Now, we know that $arctan : mathbb{R} -> (-frac{pi}{2}, frac{pi}{2})$ but if we take $a_n = frac{pi}{2}$ then $sum_{n=1}^{infty}a_n$ is divergent, but maybe there is a sequence that is bigger than $|arctan(frac{2x}{x^2+n^4})|$ but it is convergent.

Can you help me out to see how to study the convergence of this function series?

Hint: For $x ge 0$, Using AM-GM inequality we have: $x^2+n^4 ge 2xn^2implies dfrac{2x}{x^2+n^4}le dfrac{1}{n^2}implies tan^{-1}left(frac{2x}{x^2+n^4}right)le tan^{-1}left(frac{1}{n^2}right)ledfrac{1}{n^2}$ ( note: $tan^{-1}(theta) le theta, theta ge 0$ ,and $tan^{-1}(theta)$ is a increasing function of $theta$. For $x < 0$, use $tan^{-1}(-x) = -tan^{-1}(x)$.And the comparison test shows the series convergent.

We have that for any $xneq 0$

$$arctanleft(frac{2x}{x^2+n^4}right) sim frac{2x}{n^4}$$

therefore the series converges by limit comparison test with $sum frac1{n^2}$.