To find the first four elements of the sequence of partial sums of the series:

$\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}$

we first simplify the general term $\frac{1}{(2n-1)(2n+1)}$.

Step 1: Simplify the general term

We use partial fraction decomposition to express $\frac{1}{(2n-1)(2n+1)}$ as:

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

Multiplying both sides by $(2n-1)(2n+1)$, we get:

$1 = A(2n+1) + B(2n-1)$

Expanding the right-hand side:

$1 = A(2n) + A + B(2n) - B = (A + B)2n + (A - B)$

For this to hold, we need:

$A + B = 0 \quad \text{and} \quad A - B = 1$

Solving these equations, we get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$. So:

$\frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)$

Step 2: Write the partial sum

The partial sum $S_N$ is:

$S_N = \sum_{n=1}^{N} \frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \sum_{n=1}^{N} \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)$

This is a telescoping series, meaning many terms will cancel out.

Step 3: Compute the first four partial sums

For $N=1$:

$S_1 = \frac{1}{2} \left(\frac{1}{1} - \frac{1}{3}\right) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$

For $N=2$:

$S_2 = \frac{1}{2} \left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right)\right] = \frac{1}{2} \times \left(\frac{1}{1} - \frac{1}{5}\right) = \frac{1}{2} \times \frac{4}{5} = \frac{2}{5}$

For $N=3$:

$S_3 = \frac{1}{2} \left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right)\right] = \frac{1}{2} \times \left(\frac{1}{1} - \frac{1}{7}\right) = \frac{1}{2} \times \frac{6}{7} = \frac{3}{7}$

For $N=4$:

$S_4 = \frac{1}{2} \left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{9}\right)\right]$ $S_4 = \frac{1}{2} \times \left(\frac{1}{1} - \frac{1}{9}\right) = \frac{1}{2} \times \frac{8}{9} = \frac{4}{9}$

Final answer:

The first four partial sums of the series are:

$S_1 = \frac{1}{3}, \quad S_2 = \frac{2}{5}, \quad S_3 = \frac{3}{7}, \quad S_4 = \frac{4}{9}$

Would you like further details or have any questions?

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Here are five related questions:

1. How does the telescoping nature of the series simplify the calculation of partial sums?
2. What happens to the general behavior of the partial sums as $N \to \infty$?
3. Can the closed-form expression for the infinite sum of this series be found?
4. How would the partial sum change if the series involved powers of $n$ in the denominator?
5. What other techniques can be used to sum similar infinite series?

Tip: Telescoping series are highly useful because many terms cancel, leaving only a few terms to calculate, greatly simplifying the sum.