Answer
$$
\sum_{n=1}^{\infty} \frac{n}{(1+n^{2})^{p}}
$$
The given series converges if $p \gt 1$, and diverges if $0 \lt p \lt 1 $ .
Work Step by Step
$$
\sum_{n=1}^{\infty} \frac{n}{(1+n^{2})^{p}}
$$
Apply the Integral Test to the series when $p=1$
$$
\sum_{n=1}^{\infty} \frac{n}{(1+n^{2})}
$$
The function
$$
f(x)=\frac{x}{(1+x^{2})}
$$
is positive and continuous for $x\geq 1$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{1-x^{2}}{(1+x^{2})^{2}}
$$
So, $f^{'}(x) \lt 0 $ for $x\gt 1$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{1}^{\infty} \frac{x}{(1+x^{2})} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{x}{(1+x^{2})} d x \\
&=\lim\limits_{t \to \infty } [\frac{1}{2} \ln {(1+x^{2})}]_{1}^{t} \\
&=\infty \\
\end{aligned}
$$
(a) So, the series diverges when $p=1$
(b) If $p \ne 1$
Apply the Integral Test to the series when $p\ne 1$
$$
\sum_{n=1}^{\infty} \frac{n}{(1+n^{2})^{p}}
$$
The function
$$
f(x)=\frac{x}{(1+x^{2})^{p}}
$$
is positive and continuous for $x\geq 1 $ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{1-(2p-1)x^{2}}{(1+x^{2})^{p+1}}
$$
So, $f^{'}(x) \lt 0 $ for $p \gt 0 , p \ne 1$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{1}^{\infty} \frac{x}{(1+x^{2})^{p}} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{x}{(1+x^{2})^{p}} d x \\
&=\lim\limits_{t \to \infty } [\frac{1}{(x^{2}+1)^{p-1}(2-2p) } ] \\
&=\left[ \frac{1}{(x^{2}+1)^{p-1}(2-2p) } \right]_{1}^{\infty}\\
\end{aligned}
$$
we obtain that if $p \gt 1$ the integral is convergent. For $0 \lt p \lt 1 $ the integral is divergent .
Therefore, the series converges if $p \gt 1$.
