The task asks to calculate the value of the given series and find its limit:

$\sum_{k=1}^{\infty} \frac{2}{k^2 + k}$

Solution:

We begin by simplifying the term inside the sum. The general term can be rewritten as follows:

$\frac{2}{k^2 + k} = \frac{2}{k(k+1)}$

Now, using partial fraction decomposition, we can break this down:

$\frac{2}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$

Multiplying both sides by $k(k+1)$ gives:

$2 = A(k+1) + Bk$

Expanding the right-hand side:

$2 = A k + A + B k$

This implies:

$A + B = 0 \quad \text{and} \quad A = 2$

From $A = 2$, we know $B = -2$. Therefore, the partial fraction decomposition is:

$\frac{2}{k(k+1)} = \frac{2}{k} - \frac{2}{k+1}$

The Infinite Series

Now, substituting this back into the original sum:

$\sum_{k=1}^{\infty} \left( \frac{2}{k} - \frac{2}{k+1} \right)$

This is a telescoping series. In a telescoping series, most terms cancel out, leaving only a few terms from the beginning and the end. Writing out the first few terms:

$\left( \frac{2}{1} - \frac{2}{2} \right) + \left( \frac{2}{2} - \frac{2}{3} \right) + \left( \frac{2}{3} - \frac{2}{4} \right) + \dots$

Most terms cancel, and we are left with:

$\lim_{n \to \infty} \left( 2 - \frac{2}{n+1} \right)$

As $n \to \infty$, $\frac{2}{n+1} \to 0$. Therefore, the sum of the series is:

$2$

Conclusion:

The series converges to 2.

Would you like more details or have any questions about this solution?

Here are 5 related questions to deepen your understanding:

1. What are the properties of a telescoping series?
2. How does partial fraction decomposition work in other cases?
3. Can you generalize this method for other series involving rational terms?
4. How do you determine whether a series converges?
5. What is the importance of the limit in infinite series?

Tip: In telescoping series, always check which terms cancel to simplify your sum.