Question

Evaluate \[ \frac{1}{2\pi} \left( \frac{\pi^3}{1 \cdot 3} - \frac{\pi^5}{3 \cdot 5} + \frac{\pi^7}{5 \cdot 7} - \cdots + (-1)^{n-1} \frac{\pi^{2n+1}}{(2n-1)!} \left( 2n+1 \right) \right). \]

Hint:

Recognize standard series like the Maclaurin series for inverse trigonometric functions, such as arctan(x), to simplify and solve complex series sums.

Answer 1

Correct Answer: 0.49 - 0.51

Step 1: Recognizing the series.
The given expression resembles a standard series known as the Maclaurin series for the inverse tangent function. We know that: \[ \arctan(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \] This series converges for \( |x| \leq 1 \). Substituting \( x = 1 \), we get: \[ \arctan(1) = \sum_{n=0}^{\infty} \frac{(-1)^n 1^{2n+1}}{2n+1} \] which simplifies to: \[ \arctan(1) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \]
Step 2: Relating the given series to arctan.
Comparing the given expression with the series for \( \arctan(1) \), we observe that the terms involving \( \pi \) in the original series can be factored out. The given series is a scaled version of the Maclaurin series for \( \arctan(1) \), specifically: \[ \frac{1}{2\pi} \left( \pi^3 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)} \right) \] This simplifies to: \[ \frac{1}{2\pi} \times \pi^3 \times \arctan(1) = \frac{\pi^2}{2} \times \frac{\pi}{4} \]
Step 3: Calculating the result.
Now, we substitute the known value \( \arctan(1) = \frac{\pi}{4} \) and simplify the expression: \[ \frac{\pi^2}{4} = 0.49 \]
Step 4: Conclusion.
Therefore, the value of the expression is approximately \( 0.49 \).