Question

The sum of the infinite series \[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \pi^{2n+1}}{2^{2n+1} (2n)!} \] is equal to

• A. \( -\pi \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( -\frac{\pi}{4} \)

Hint:

When dealing with series, identify patterns that resemble known expansions (like sine, cosine, etc.) to simplify the evaluation.

Answer 1

The Correct Option is C

Step 1: Recognize the series form.
The given series is \[ S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \pi^{2n+1}}{2^{2n+1} (2n)!} \] This resembles the series expansion for the sine function, which is given by: \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \] Thus, the given series is a modification of the sine series. Step 2: Express the series in terms of sine.
We can rewrite the series by factoring out \( \pi \) from the powers of \( \pi^{2n+1} \) and simplifying the factorials: \[ S = \pi \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \pi^{2n}}{2^{2n+1} (2n)!} \] This can be rewritten as a modified sine series form. Step 3: Simplify and match with known series.
Recognizing the structure of the modified sine series, we find that the sum converges to: \[ S = \frac{\pi}{2} \] Final Answer: \[ \boxed{\frac{\pi}{2}} \]