Answer
$$
\sum_{n=1}^{\infty} \frac{1}{n^{5}}
$$
Apply the Integral Test to the given series we obtain
$$
\int_{1}^{\infty} \frac{1}{x^{5}} d x= \frac{1}{4 }
$$
So, the series converges
Work Step by Step
$$
\sum_{n=1}^{\infty} \frac{1}{n^{5}}
$$
Apply the Integral Test to the series
$$
\sum_{n=1}^{\infty} \frac{1}{n^{5}}
$$
The function
$$
f(x)=\frac{1}{x^{5}}
$$
is positive and continuous for $x\geq 1$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{-5}{x^{6}}
$$
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
$$
\begin{aligned}
\int_{1}^{\infty} \frac{1}{x^{5}} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{1}{x^{5}} d x \\
&=\lim\limits_{t \to \infty } [\frac{-1}{4x^{4} } ]_{1}^{t} \\
&=[\frac{-1}{4x^{4} }]_{1}^{\infty} \\
&=\lim\limits_{t \to \infty }[\frac{-1}{4t^{4} }] +[\frac{1}{4 }]\\
&=0+\frac{1}{4 } \\
&=\frac{1}{4 }
\end{aligned}
$$
So, the series converges