The provided image presents the following series:

$\sum_{n=1}^{\infty} \left( \frac{1}{n^3} - \frac{1}{(n+1)^3} \right)$

This is a telescoping series, where terms from successive summations cancel each other out, leaving only a few terms in the final sum.

Step-by-step approach:

1. Express the partial sum: Consider the partial sum up to a finite value $N$:

   $S_N = \sum_{n=1}^{N} \left( \frac{1}{n^3} - \frac{1}{(n+1)^3} \right)$

   By expanding the terms for $n = 1$ to $n = N$, we can see that the majority of terms cancel out (since for each $n$, $\frac{1}{(n+1)^3}$ cancels with $\frac{1}{n^3}$ from the next term).

2. Cancellation of terms: The first term from $n = 1$ is $\frac{1}{1^3} = 1$, and the last term after cancellation is $\frac{1}{(N+1)^3}$. Therefore, the partial sum simplifies to:

   $S_N = 1 - \frac{1}{(N+1)^3}$

3. Take the limit as $N \to \infty$: Now, we compute the limit of $S_N$ as $N$ approaches infinity:

   $\lim_{N \to \infty} S_N = \lim_{N \to \infty} \left( 1 - \frac{1}{(N+1)^3} \right) = 1 - 0 = 1$

Thus, the sum of the series is:

$\boxed{1}$

Let me know if you would like more details or have any questions.

Here are 5 related questions for further exploration:

1. How does a telescoping series differ from a geometric series?
2. What conditions ensure the convergence of a telescoping series?
3. Can all series be simplified into a telescoping form? Why or why not?
4. What is an example of a divergent series, and how does it compare to this one?
5. How does the sum of an infinite geometric series compare to that of a telescoping series?

Tip: When evaluating telescoping series, always look for cancellations between consecutive terms, as this often leads to significant simplification!