Find the first four elements of the sequence of partial sum $$ \sum $$$$ \frac{1}{(2n-1)(2n+1)} $$, n=1 to infinity 

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.

Find the first four elements of the sequence of partial sums $$ \sum $$ $$ \frac{1}{(2n-1)(2n+1)} $$, n=1 to infinity.

Discover how to find the first four elements of the partial sums sequence for the infinite series involving $ \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} $. This problem explores telescoping series, partial sums, and their simplification using partial fractions. Suitable for students in Grades 11-12 and early college learners.

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