The series can be written as [latex]\displaystyle\sum_{n=1}^{\infty} \frac{4(-1)^{n+1}}{2n-1}[/latex].

This is an alternating series with [latex]a_n = \frac{4}{2n-1}[/latex].

To apply the alternating series test, we need to verify:

1. [latex]a_n > 0[/latex] for all [latex]n \geq 1[/latex]
2. [latex]a_{n+1} \leq a_n[/latex] for all [latex]n \geq 1[/latex] (decreasing)
3. [latex]\lim_{n \to \infty} a_n = 0[/latex]

Verification:

1. [latex]a_n = \frac{4}{2n-1} > 0[/latex] since [latex]2n-1 > 0[/latex] for all [latex]n \geq 1[/latex]
2. We need [latex]a_{n+1} \leq a_n[/latex], which means: [latex]\frac{4}{2(n+1)-1} \leq \frac{4}{2n-1}[/latex] [latex]\frac{4}{2n+1} \leq \frac{4}{2n-1}[/latex] Since [latex]2n+1 > 2n-1[/latex] for [latex]n \geq 1[/latex], we have [latex]\frac{1}{2n+1} < \frac{1}{2n-1}[/latex], so the sequence is decreasing.
3. [latex]\lim_{n \to \infty} \frac{4}{2n-1} = 0[/latex]

Since all conditions are satisfied, the series converges by the alternating series test.

[latex]\begin{array}{rcl} S_{10} & = & 4\left(1 - \tfrac{1}{3} + \tfrac{1}{5} - \tfrac{1}{7} + \tfrac{1}{9} - \tfrac{1}{11} + \tfrac{1}{13} - \tfrac{1}{15} + \tfrac{1}{17} - \tfrac{1}{19}\right) \\ & = & 4(0.760460) = 3.041840 \\[6pt] S_{20} & = & 4\sum_{n=1}^{20} \frac{(-1)^{\,n+1}}{2n-1} = 4(0.772840) = 3.091361 \\[6pt] S_{50} & = & 4\sum_{n=1}^{50} \frac{(-1)^{\,n+1}}{2n-1} = 4(0.780398) = 3.121593 \\[6pt] S_{100} & = & 4\sum_{n=1}^{100} \frac{(-1)^{\,n+1}}{2n-1} = 4(0.782898) = 3.131593 \\ \end{array}[/latex]

For an alternating series, the remainder estimate states that [latex]|R_n| \leq a_{n+1}[/latex].

Therefore: [latex]|R_n| \leq \frac{4}{2(n+1)-1} = \frac{4}{2n+1}[/latex]

For specific values:

• [latex]|R_{10}| \leq \frac{4}{21} \approx 0.190[/latex]
• [latex]|R_{20}| \leq \frac{4}{41} \approx 0.098[/latex]
• [latex]|R_{50}| \leq \frac{4}{101} \approx 0.040[/latex]
• [latex]|R_{100}| \leq \frac{4}{201} \approx 0.020[/latex]

We need [latex]\frac{4}{2N+1} < 0.01[/latex]

[latex]\frac{4}{2N+1} < 0.01[/latex] [latex]4 < 0.01(2N+1)[/latex] [latex]4 < 0.02N + 0.01[/latex] [latex]3.99 < 0.02N[/latex] [latex]N > 199.5[/latex]

Therefore, [latex]N = 200[/latex].

[latex]S_{200} = 4\sum_{n=1}^{200} \frac{(-1)^{n+1}}{2n-1} \approx 3.136593[/latex]

Let [latex]a_n = \frac{(2n)!}{2^{4n+1}(n!)^2(2n+1)}[/latex].

Using the ratio test:
[latex]\begin{array}{rcl} \dfrac{a_{n+1}}{a_n} & = & \dfrac{(2n+2)!}{2^{4n+5}((n+1)!)^2(2n+3)} \cdot \dfrac{2^{4n+1}(n!)^2(2n+1)}{(2n)!} \\ & = & \dfrac{(2n+2)(2n+1)(2n)!}{2^4(n+1)^2(n!)^2(2n+3)} \cdot \dfrac{2^{4n+1}(n!)^2(2n+1)}{2^{4n+1}(2n)!} \\ & = & \dfrac{(2n+2)(2n+1)^2}{16(n+1)^2(2n+3)} \\ & = & \dfrac{2(n+1)(2n+1)^2}{16(n+1)^2(2n+3)} \\ & = & \dfrac{(2n+1)^2}{8(n+1)(2n+3)} \end{array}[/latex]

As [latex]n \to \infty[/latex]: [latex]\lim_{n \to \infty} \frac{(2n+1)^2}{8(n+1)(2n+3)} = \lim_{n \to \infty} \frac{4n^2}{16n^2} = \frac{1}{4} < 1[/latex]

Since the ratio test limit is less than 1, the series converges.

[latex]S_5 = 6\sum_{n=0}^{5} \frac{(2n)!}{2^{4n+1}(n!)^2(2n+1)}[/latex]

Computing each term:

• [latex]n=0: \frac{1!}{2^1 \cdot 1^2 \cdot 1} = \frac{1}{2}[/latex]
• [latex]n=1: \frac{2!}{2^5 \cdot 1^2 \cdot 3} = \frac{2}{96} = \frac{1}{48}[/latex]
• [latex]n=2: \frac{4!}{2^9 \cdot 4 \cdot 5} = \frac{24}{10240} = \frac{3}{1280}[/latex]
• [latex]n=3: \frac{6!}{2^{13} \cdot 36 \cdot 7} = \frac{720}{2064384} \approx 0.000349[/latex]
• [latex]n=4: \frac{8!}{2^{17} \cdot 576 \cdot 9} \approx 0.0000596[/latex]
• [latex]n=5: \frac{10!}{2^{21} \cdot 14400 \cdot 11} \approx 0.0000109[/latex]

[latex]S_5 = 6(0.5 + 0.02083 + 0.00234 + 0.000349 + 0.0000596 + 0.0000109) \approx 3.14159[/latex]

[latex]S_{10} \approx 3.141592653[/latex]

[latex]S_{20} \approx 3.141592653589793[/latex]

[latex]\pi = 3.141592653589793...[/latex]

• [latex]S_5 \approx 3.14159[/latex]: Correct to 5 decimal places
• [latex]S_{10} \approx 3.141592653[/latex]: Correct to 9 decimal places
• [latex]S_{20} \approx 3.141592653589793[/latex]: Correct to 15 decimal places

Newton’s series converges much faster than the Gregory-Leibniz series, gaining approximately 6 additional correct decimal places with each 10 additional terms.

Let [latex]a_n = \frac{(4n)!(1103+26390n)}{(n!)^4 \cdot 396^{4n}}[/latex].

Using the ratio test:

[latex]\begin{array}{rcl} \dfrac{a_{n+1}}{a_n} & = & \dfrac{(4n+4)!\,(1103+26390(n+1))}{((n+1)!)^4 \cdot 396^{4(n+1)}} \cdot \dfrac{(n!)^4 \cdot 396^{4n}}{(4n)!\,(1103+26390n)} \\ & = & \dfrac{(4n+4)(4n+3)(4n+2)(4n+1)(1103+26390n+26390)}{(n+1)^4 \cdot 396^4} \cdot \dfrac{1}{1103+26390n} \end{array}[/latex]

As [latex]n \to \infty[/latex], the dominant terms give:

[latex]\begin{array}{rcl} \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} & = & \lim_{n \to \infty} \dfrac{256n^4 \cdot 26390n}{n^4 \cdot 396^4 \cdot 26390n} \\ & = & \dfrac{256}{396^4} \\ & = & \dfrac{256}{24596196096} < 1 \\\end{array}[/latex]

Since the ratio is less than 1, the series converges by the ratio test.

First term ([latex]n=0[/latex]):
[latex]a_0 = \frac{0! \cdot 1103}{(0!)^4 \cdot 396^0} = 1103[/latex]

[latex]\frac{1}{\pi} \approx \frac{\sqrt{8}}{9801} \cdot 1103 = \frac{2\sqrt{2} \cdot 1103}{9801} \approx 0.318309886183791[/latex]

Therefore: [latex]\pi \approx \frac{1}{0.318309886183791} \approx 3.141592653589793[/latex]

Comparing with [latex]\pi = 3.141592653589793...[/latex], the first term alone gives [latex]\pi[/latex] correct to 15 decimal places!

Adding the second term ([latex]n=1[/latex]):
[latex]a_1 = \frac{4! \cdot (1103+26390)}{(1!)^4 \cdot 396^4} = \frac{24 \cdot 27493}{396^4} \approx 2.68 \times 10^{-8}[/latex]

[latex]\frac{1}{\pi} \approx \frac{\sqrt{8}}{9801}(1103 + 2.68 \times 10^{-8})[/latex]

This gives [latex]\pi[/latex] accurate to approximately 23 decimal places.