# How do you use the Nth term test on the infinite series sum_(n=1)^ooe^2/n^3 ?

By pulling ${e}^{2}$ out of the summation,
${\sum}_{n = 1}^{\infty} {e}^{2} / {n}^{3} = {e}^{2} {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 3$,
which is a p-series with $p = 3$.
Since $p > 1$, we can conclude that the series is convergent by P-Series Test.