Let’s break down and answer each part of the worksheet: --- ### **Leibniz Series for Pi** \[ \pi = \sum_{n=1}^{\infty} (-1)^{n+1} \cdot \frac{4}{2n-1} \] #### **A. Write the first 10 terms of the sequence.** Plug in \( n = 1 \) to \( 10 \): 1. \( n = 1: \quad 4/1 = 4 \) 2. \( n = 2: \quad -4/3 \) 3. \( n = 3: \quad 4/5 \) 4. \( n = 4: \quad -4/7 \) 5. \( n = 5: \quad 4/9 \) 6. \( n = 6: \quad -4/11 \) 7. \( n = 7: \quad 4/13 \) 8. \( n = 8: \quad -4/15 \) 9. \( n = 9: \quad 4/17 \) 10. \( n = 10: \quad -4/19 \) Sequence: \( 4, -\frac{4}{3}, \frac{4}{5}, -\frac{4}{7}, \frac{4}{9}, -\frac{4}{11}, \frac{4}{13}, -\frac{4}{15}, \frac{4}{17}, -\frac{4}{19} \) --- #### **B. Find partial sums using a calculator.** \[ S_{n} = \sum_{k=1}^{n} (-1)^{k+1} \frac{4}{2k-1} \] * \( S_{10} \approx 3.04184 \) * \( S_{100} \approx 3.13159 \) * \( S_{1000} \approx 3.14059 \) (Rounded to five decimal places) --- #### **C. Show that the series converges.** The Leibniz series is an **alternating series** (terms alternate in sign), and the absolute value of its terms decreases monotonically to zero. By the **Alternating Series Test**, the series therefore converges. --- ### **Newton's Series for Pi** \[ \pi = \sum_{n=0}^{\infty} \frac{6(2n)!}{2^{4n+1}(n!)^2(2n+1)} \] #### **E. Write the first 5 terms of the sequence.** Let’s compute for \( n = 0 \) to \( 4 \): - \( n = 0: \frac{6 \cdot 1}{2^1 \cdot 1^2 \cdot 1} = \frac{6}{2 \cdot 1 \cdot 1} = 3 \) - \( n = 1: \frac{6 \cdot 2!}{2^5 \cdot 1^2 \cdot 3} = \frac{6 \cdot 2}{32 \cdot 1 \cdot 3} = \frac{12}{96} = 0.125 \) - \( n = 2: \frac{6 \cdot 4!}{2^9 \cdot 2^2 \cdot 5} = \frac{6 \cdot 24}{512 \cdot 4 \cdot 5} = \frac{144}{10240} = 0.01406 \) - \( n = 3: \frac{6 \cdot 6!}{2^{13} \cdot 6^2 \cdot 7} = \frac{6 \cdot 720}{8192 \cdot 36 \cdot 7} = \frac{4320}{2064384} = 0.00209 \) - \( n = 4: \frac{6 \cdot 8!}{2^{17} \cdot 24^2 \cdot 9} = \frac{6 \cdot 40320}{131072 \cdot 576 \cdot 9} = \frac{241920}{68024448} \approx 0.0000036 \) So, the first five terms are: \( 3, 0.125, 0.01406, 0.00209, 0.0000036 \) --- #### **F. Find partial sums using a calculator.** - \( S_3 = 3 + 0.125 + 0.01406 + 0.00209 = 3.14115 \) - \( S_5 \) (add next two terms) \( \approx 3.14159 \) - \( S_{10} \approx 3.14159 \) --- #### **G. Which formula for pi do you think is better?** **Newton’s formula is better** because it converges to pi much faster than the Leibniz series. This means you need far fewer terms to get an accurate value of pi, making it more efficient and practical for calculations.

Certainly! Here is a detailed step-by-step solution with all calculations and final answers: --- ### **Part A: First 10 Terms of Leibniz Series for Pi** The series is: \[ \pi = \sum_{n=1}^\infty (-1)^{n+1} \frac{4}{2n-1} \] **Calculations for the first 10 terms:** | \( n \) | Term \( T_n = (-1)^{n+1} \frac{4}{2n-1} \) | Calculation | Decimal approximation | |---------|---------------------------------------------|------------------------------|------------------------| | 1 | \( (+1) \times \frac{4}{1} \) | \( 4/1 \) | **4.00000** | | 2 | \( (-1) \times \frac{4}{3} \) | \( -4/3 \) | **-1.33333** | | 3 | \( (+1) \times \frac{4}{5} \) | \( 4/5 \) | **0.80000** | | 4 | \( (-1) \times \frac{4}{7} \) | \( -4/7 \) | **-0.57143** | | 5 | \( (+1) \times \frac{4}{9} \) | \( 4/9 \) | **0.44444** | | 6 | \( (-1) \times \frac{4}{11} \) | \( -4/11 \) | **-0.36364** | | 7 | \( (+1) \times \frac{4}{13} \) | \( 4/13 \) | **0.30769** | | 8 | \( (-1) \times \frac{4}{15} \) | \( -4/15 \) | **-0.26667** | | 9 | \( (+1) \times \frac{4}{17} \) | \( 4/17 \) | **0.23529** | | 10 | \( (-1) \times \frac{4}{19} \) | \( -4/19 \) | **-0.21053** | --- ### **Part B: Partial Sums for the First 10, 100, and 1000 Terms** **Calculations:** - **Sum of first 10 terms \( S_{10} \):** \[ S_{10} = 4 - 1.33333 + 0.80000 - 0.57143 + 0.44444 - 0.36364 + 0.30769 - 0.26667 + 0.23529 - 0.21053 \] Step-by-step: 1. \( 4 - 1.33333 = 2.66667 \) 2. \( 2.66667 + 0.80000 = 3.46667 \) 3. \( 3.46667 - 0.57143 = 2.89524 \) 4. \( 2.89524 + 0.44444 = 3.33968 \) 5. \( 3.33968 - 0.36364 = 2.97604 \) 6. \( 2.97604 + 0.30769 = 3.28373 \) 7. \( 3.28373 - 0.26667 = 3.01706 \) 8. \( 3.01706 + 0.23529 = 3.25235 \) 9. \( 3.25235 - 0.21053 = 3.04182 \) **Rounded to five decimal places:** \[ \boxed{S_{10} \approx 3.04184} \] --- - **Sum of first 100 terms \( S_{100} \):** Using calculator (or software), the partial sum is approximately: \[ \boxed{S_{100} \approx 3.13159} \] - **Sum of first 1000 terms \( S_{1000} \):** Similarly computed: \[ \boxed{S_{1000} \approx 3.14059} \] --- ### **Part C: Showing the Series Converges** This series is an **alternating series** where: - The absolute value of the terms decreases monotonically: \( \frac{4}{2n-1} \) gets smaller as \( n \) increases. - The limit of the terms as \( n \to \infty \) is zero. According to the **Alternating Series Test**, these conditions confirm the series converges. --- ### **Part E: First 5 Terms of Newton's Series for Pi** The formula: \[ \pi = \sum_{n=0}^\infty \frac{6(2n)!}{2^{4n+1} (n!)^2 (2n+1)} \] Calculate for \( n=0 \) to \( 4 \): | \( n \) | Term \( T_n \) | Calculation | Decimal approximation | |---------|----------------|------------------------------|------------------------| | 0 | \(\frac{6 \times 0!}{2^{1} \times (0!)^2 \times 1}\) | \(\frac{6 \times 1}{2 \times 1 \times 1}\) | \(6/2 = 3.00000\) | | 1 | \(\frac{6 \times 2!}{2^{5} \times (1!)^2 \times 3}\) | \(\frac{6 \times 2}{32 \times 1 \times 3}\) | \(\frac{12}{96} = 0.12500\) | | 2 | \(\frac{6 \times 4!}{2^{9} \times (2!)^2 \times 5}\) | \(\frac{6 \times 24}{512 \times 4 \times 5}\) | \(\frac{144}{10240} \approx 0.01406\) | | 3 | \(\frac{6 \times 6!}{2^{13} \times (3!)^2 \times 7}\) | \(\frac{6 \times 720}{8192 \times 36 \times 7}\) | \(\frac{4320}{2064384} \approx 0.00209\) | | 4 | \(\frac{6 \times 8!}{2^{17} \times (4!)^2 \times 9}\) | \(\frac{6 \times 40320}{131072 \times 576 \times 9}\) | \(\frac{241920}{68024448} \approx 0.0000036\) | **First five terms:** \[ \boxed{ 3, \quad 0.125, \quad 0.01406, \quad 0.00209, \quad 0.0000036 } \] --- ### **Part F: Partial Sums for Newton's Series** Sum of first 3 terms: \[ S_3 = 3 + 0.125 + 0.01406 = 3.13906 \] Adding the 4th term: \[ S_4 = 3.13906 + 0.00209 \approx 3.14115 \] Adding the 5th term: \[ S_5 \approx 3.14115 + 0.0000036 \approx 3.14116 \] **Note:** As the sum approaches more terms, it converges rapidly to \(\pi \approx 3.14159\). --- ### **Part G: Which formula for pi is better?** **Answer:** *Newton's series converges much faster than Leibniz's series.* This means fewer terms are needed to approximate pi accurately. For example, after only 5 terms, Newton's series already gets close to 3.14159, whereas Leibniz's series requires many more terms to reach similar accuracy. Therefore, Newton's formula is more efficient for calculating pi. --- **Summary of Final Answers:** | Part | Final Answer | |---------|--------------| | A | First 10 terms: \( 4, -\frac{4}{3}, \frac{4}{5}, -\frac{4}{7}, \frac{4}{9}, -\frac{4}{11}, \frac{4}{13}, -\frac{4}{15}, \frac{4}{17}, -\frac{4}{19} \) | | B | Partial sums: \( S_{10} \approx 3.04184 \), \( S_{100} \approx 3.13159 \), \( S_{1000} \approx 3.14059 \) | | C | The Leibniz series converges by the Alternating Series Test because the terms decrease monotonically to zero. | | E | First 5 terms of Newton's series: 3, 0.125, 0.01406, 0.00209, 0.0000036 | | F | Partial sums: \( S_3 \approx 3.13906 \), \( S_4 \approx 3.14115 \), \( S_5 \approx 3.14116 \) | | G | Newton's series is better because it converges faster, requiring fewer terms to approximate pi accurately. | If you need any further clarification, feel free to ask!