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The image you uploaded contains two mathematical problems related to series. Let's go through each one to determine whether the series is convergent or divergent, and if convergent, find its sum.

### Problem (a)
The first problem asks you to evaluate the series:
\[
\sum_{k=1}^{\infty} \left(\frac{1}{k^2 + 1} - \frac{1}{k^2 + 2}\right)
\]

#### Solution for (a):
Notice that the series is telescoping. Let's write out the first few terms to see this more clearly:
\[
\left(\frac{1}{1^2+1} - \frac{1}{1^2+2}\right) + \left(\frac{1}{2^2+1} - \frac{1}{2^2+2}\right) + \left(\frac{1}{3^2+1} - \frac{1}{3^2+2}\right) + \cdots
\]
\[
= \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \left(\frac{1}{10} - \frac{1}{11}\right) + \cdots
\]

Each pair of terms shows that most terms will cancel out in a telescoping fashion, leaving us with only the first and the very last terms from each part of the series. 

So, the sum of the series is:
\[
\lim_{n \to \infty} \left( \frac{1}{2} - \frac{1}{n^2+2} \right) = \frac{1}{2} - 0 = \frac{1}{2}
\]
Thus, the series converges, and the sum is \(\frac{1}{2}\).

### Problem (b)
The second problem asks you to evaluate the series:
\[
\sum_{k=2}^{\infty} \frac{k}{(k-1)!}
\]

#### Solution for (b):
This series can be simplified by noticing that \(k = (k-1) + 1\). Thus:
\[
\frac{k}{(k-1)!} = \frac{(k-1) + 1}{(k-1)!} = \frac{(k-1)}{(k-1)!} + \frac{1}{(k-1)!}
\]
\[
= \frac{(k-1)!}{(k-1)!} + \frac{1}{(k-1)!} = 1 + \frac{1}{(k-1)!}
\]

Therefore, the series becomes:
\[
\sum_{k=2}^{\infty} \left(1 + \frac{1}{(k-1)!}\right)
\]

The series can now be split into two simpler series:
\[
\sum_{k=2}^{\infty} 1 + \sum_{k=2}^{\infty} \frac{1}{(k-1)!}
\]

The first series, \(\sum_{k=2}^{\infty} 1\), is a divergent series (it is simply an infinite sum of 1's).

The second series, \(\sum_{k=2}^{\infty} \frac{1}{(k-1)!}\), is actually convergent and related to the exponential series. 

However, since the first series diverges, the overall series diverges.

### Conclusion:
- **(a)** The series converges with a sum of \(\frac{1}{2}\).
- **(b)** The series diverges.

Would you like any additional details or explanations?

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**Related Questions:**
1. How can you determine if a series is telescoping?
2. What are some examples of other telescoping series?
3. How do you determine convergence for factorial-related series?
4. What is the difference between an absolutely convergent series and a conditionally convergent series?
5. Can you find another example of a series that diverges similarly to part (b)?

**Tip:** When working with series, always check if the terms simplify or telescope—this often reveals the series' convergence properties.

Determine whether the given series is convergent or divergent. If convergent, find its sum. (a) Evaluate ∑(k=1 to ∞) [1/(k^2 + 1) - 1/(k^2 + 2)]. (b) Evaluate ∑(k=2 to ∞) [k / ((k-1)!)]

Determine the convergence of series involving telescoping and factorial simplifications. This solution provides a step-by-step evaluation of whether the series converges or diverges and finds the sum for the convergent series.

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