Answer
Converges
Work Step by Step
We are given the series as: $\Sigma_{n=1}^{\infty} \dfrac{1}{n^3+1}$
We can see that the degree of the denominator is equal to $k=3$ and the degree of the numerator is equal to $j=0$.
This implies that $j \lt k$
Also, the condition $ j=0 \lt k-1=21$ has been satisfied.
Hence, we can conclude that the given series converges.
