The given series to test for convergence is:

[ \sum \left( \sqrt[3]{n^3 + 1} - \sqrt[3]{n^3} \right) ]

To determine the convergence of this series, we can start by simplifying the general term:

Step 1: Simplification of the General Term

The term inside the summation can be simplified using the binomial approximation for cube roots:

[ \sqrt[3]{n^3 + 1} = \sqrt[3]{n^3 \left( 1 + \frac{1}{n^3} \right)} = n \cdot \sqrt[3]{1 + \frac{1}{n^3}} ]

For large $n$, we can approximate:

$\sqrt[3]{1 + \frac{1}{n^3}} \approx 1 + \frac{1}{3n^3}$

So,

$\sqrt[3]{n^3 + 1} \approx n \left(1 + \frac{1}{3n^3}\right) = n + \frac{1}{3n^2}$

Therefore,

[ \sqrt[3]{n^3 + 1} - \sqrt[3]{n^3} \approx \left(n + \frac{1}{3n^2}\right) - n = \frac{1}{3n^2} ]

Step 2: Behavior of the Series

The series now becomes:

$\sum \frac{1}{3n^2}$

This is a p-series with $p = 2$, which is known to converge because $p > 1$.

Conclusion

Since the series behaves asymptotically like $\sum \frac{1}{3n^2}$, which converges, the original series:

[ \sum \left( \sqrt[3]{n^3 + 1} - \sqrt[3]{n^3} \right) ]

also converges.

Do you have any questions or would you like further details on any step?

Here are 5 related questions you might find interesting:

1. How does the binomial approximation apply to other roots, such as square roots?
2. Can the Ratio Test be applied to this series, and if so, what does it reveal?
3. What is the behavior of the series if we modify the exponent from 3 to another integer?
4. How does the p-series test work for determining the convergence of series?
5. What happens if we consider the series sum from a different starting index?

Tip: When testing series for convergence, simplifying the general term can often reveal the nature of the series, making it easier to apply known convergence tests.