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$$  $$Can you calculate the sum of the series of question 36?

Let's analyze and solve the series given in question 36. The series is:

\[ \sum_{p=2}^{\infty} \frac{1}{\sqrt{p-1} + \sqrt{p+1}} \]

To simplify the series, we can start by rationalizing the denominator. This involves multiplying the numerator and denominator by the conjugate of the denominator:

\[ \frac{1}{\sqrt{p-1} + \sqrt{p+1}} \times \frac{\sqrt{p-1} - \sqrt{p+1}}{\sqrt{p-1} - \sqrt{p+1}} = \frac{\sqrt{p-1} - \sqrt{p+1}}{(\sqrt{p-1} + \sqrt{p+1})(\sqrt{p-1} - \sqrt{p+1})} \]

The denominator simplifies using the difference of squares:

\[ (\sqrt{p-1})^2 - (\sqrt{p+1})^2 = (p-1) - (p+1) = p - 1 - p - 1 = -2 \]

So the expression becomes:

\[ \frac{\sqrt{p-1} - \sqrt{p+1}}{-2} = -\frac{\sqrt{p-1} - \sqrt{p+1}}{2} \]

Therefore, the series can be written as:

\[ \sum_{p=2}^{\infty} -\frac{\sqrt{p-1} - \sqrt{p+1}}{2} \]

This is a telescoping series. To see why, let's write out the first few terms:

\[ -\frac{\sqrt{1} - \sqrt{3}}{2}, -\frac{\sqrt{2} - \sqrt{4}}{2}, -\frac{\sqrt{3} - \sqrt{5}}{2}, \ldots \]

Notice that most terms will cancel out:

\[ -\frac{1 - \sqrt{3}}{2} - \frac{\sqrt{2} - 2}{2} - \frac{\sqrt{3} - \sqrt{5}}{2} \ldots \]

The general cancellation pattern looks like this:

\[ -\frac{1}{2} + \frac{\sqrt{p+1}}{2} \]

Since the terms cancel out, only the first and last terms remain after cancellation. As $ p \to \infty $, $\sqrt{p+1} \to \infty$. Therefore, the series does not converge because the terms do not sum to a finite number.

Thus, the series diverges.

Do you have any questions or need further details?

Here are some related questions that might interest you:
1. Can you explain the concept of a telescoping series?
2. How do you determine if a series converges or diverges?
3. What is the difference of squares formula and how is it used in series?
4. How do you rationalize the denominator?
5. Can you provide an example of a convergent telescoping series?
6. What are the common tests for convergence of an infinite series?
7. How does the behavior of the terms in a series affect its convergence?
8. What are some practical applications of series in real life?

**Tip:** When working with series, identifying patterns such as telescoping can greatly simplify the process of determining convergence or divergence.

Explore how to calculate and analyze the sum of the series in Question 36, focusing on telescoping series and their convergence properties. Detailed step-by-step solution provided.

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