Convergence of: $\sum_{n=1}^{\infty}\frac{2nx^2}{1+n^2 x^4}$

Question

I have to calculate the convergence of this serie:

$$\sum_{n=1}^{\infty}\frac{2nx^2}{1+n^2 x^4} \space\text{being} \space x\in [1,2]$$

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I have calculated the derivate and i know that $f_n$ has a maximum in $x=\frac{1}{\sqrt n}$, so I can know that $f_n$ is descending in $[1,2]$

It's correct to say that the serie is divergence because

$$\sum_{n=1}^{\infty} f_n(2)\leq \sum_{n=1}^{\infty}f_n(x)$$ and $\sum_{n=1}^{\infty} f_n(2)$ is divergence?

Answer 1

Hint: Use the integral criterion:

\begin{equation} \int\frac{2nx^2}{1+n^2x^4}dn=\frac{\ln(1+n^2x^4)}{x^2}+constant \end{equation}

Answer 2

In your interval you have

$$ \frac{2nx^2}{1+n^2 x^4} > \frac{1}{2} \frac{2nx^2}{n^2 x^4} = \frac{x^{-2}}{n} $$

And the latter series is diverging.

Answer 3

$$\frac{2nx^2}{1+n^2 x^4} = \frac{2nx^2}{n^2 x^4 \left( 1+ \frac{1}{n^2 x^4}\right)} = \frac{1}{n } \cdot \frac{2}{x^2\left( 1+ \frac{1}{n^2 x^4}\right)} \sim \frac{1}{n } \cdot \frac{2}{x^2}$$