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