Answer
Converges
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{1}{n^{4}}$$
Since $f(n)= \frac{1}{n^{4}}$, is positive, decreasing and continuous for $n\geq 1$, then by using the integral test
\begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{R} \frac{1}{n^{4}} d n \\ &=\left.\lim _{t \rightarrow \infty} \frac{-1}{3 n^{3}}\right|_{1} ^{t} \\ &=\lim _{t \rightarrow \infty}\left(\frac{1}{3}-\frac{1}{3 t^{3}}\right) \\ &=\frac{1}{3} \end{aligned}
Thus the series converges.
