Answer
$$
\sum_{n=1}^{\infty} \frac{n}{n^{4}+2n^{2}+1}
$$
Apply the Integral Test to the given series we obtain
$$
\int_{1}^{\infty}\frac{x}{(x^{2}+1)^{2}} d x =\frac{1}{4}.
$$
So, the series diverges
Work Step by Step
$$
\sum_{n=1}^{\infty} \frac{n}{n^{4}+2n^{2}+1}
$$
Apply the Integral Test to the series
$$
\sum_{n=1}^{\infty} \frac{n}{n^{4}+2n^{2}+1}
$$
The function
$$
f(x)=\frac{x}{x^{4}+2x^{2}+1}=\frac{x}{(x^{2}+1)^{2}}
$$
is positive and continuous for $x\geq 1$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{-(3x^{2}-1)} {(x^{2}+1)^{2}}
$$
So, $f^{'}(x) \lt 0 $ for $x \geq 1$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{1}^{\infty}\frac{x}{(x^{2}+1)^{2}} d x &= \lim\limits_{t \to \infty } \frac{1}{2}\int_{1}^{t } \frac{2x}{(x^{2}+1)^{2}} d x \\
& \quad\quad\quad \left[\text { use integration by substituting }\right] \\
&\quad\quad\quad \left[ u=x^{2}+1, \quad\quad d u= 2x dx \right] , \text { then }\\
&= \lim\limits_{t \to \infty } \frac{1}{2}\int_{1}^{t } \frac{du}{u^{2}} \\
&=\lim\limits_{t \to \infty } \left[ \frac{-1}{2 u } \right] _{1}^{t} \\
&=\lim\limits_{t \to \infty } \left[ \frac{-1}{2 (x^{2}+1) }\right] _{1}^{t} \\
&= \left[ \lim\limits_{t \to \infty }\frac{-1}{2 (t^{2}+1) }+\frac{1}{4} \right] \\
&= \left[ 0+\frac{1}{4} \right] \\
&=\frac{1}{4}.
\end{aligned}
$$
So, the series converges.