Answer
$$
\sum_{n=1}^{\infty} \frac{n}{n^{4}+1}
$$
Apply the Integral Test to the given series we obtain
$$
\int_{1}^{\infty}\frac{x}{x^{4}+1} d x =\frac{\pi}{8}.
$$
So, the series converges
Work Step by Step
$$
\sum_{n=1}^{\infty} \frac{n}{n^{4}+1}
$$
Apply the Integral Test to the series
$$
\sum_{n=1}^{\infty} \frac{n}{n^{4}+1}
$$
The function
$$
f(x)=\frac{x}{x^{4}+1}
$$
is positive and continuous for $x\geq 1$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{1-3x^{4}} {(x^{4}+1)^{2}}
$$
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
$$
\begin{aligned}
\int_{1}^{\infty}\frac{x}{x^{4}+1} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{x}{x^{4}+1} d x \\
& \quad\quad\quad \left[\text { use integration by substituting }\right] \\
&\quad\quad\quad \left[ u=x^{2}, \quad\quad d u= 2x dx \right] , \text { then }\\
&= \lim\limits_{t \to \infty } \frac{1}{2}\int_{1}^{t } \frac{2x}{x^{4}+1} d x \\
&= \lim\limits_{t \to \infty } \frac{1}{2}\int_{1}^{t } \frac{du}{u^{2}+1} \\
&=\lim\limits_{t \to \infty } \left[ \frac{1}{2 }\arctan u \right] _{1}^{t} \\
&=\lim\limits_{t \to \infty } \left[ \frac{1}{2 }\arctan x^{2} \right] _{1}^{t} \\
&= \frac{1}{2 } \left[ \lim\limits_{t \to \infty } \arctan t^{2}-\frac{\pi}{4} \right] \\
&= \frac{1}{2 } \left[ \frac{\pi}{2}-\frac{\pi}{4} \right] \\
&=\frac{\pi}{8}.
\end{aligned}
$$
So, the series converges.