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