which is correct, representing sin(\frac{(2k+1)pi}{2}) as \sum_{k=0}^{\infty} \frac{(-1)^{2k+1}}{(2k+1)^2} (-1)^k or \sum_{k=0}^{\infty} \frac{(-1)^{2k+1}}{(2k+1)^2} (-1)^k+1

Let's analyze both representations and determine which one is correct.

### Expression 1:
\[
\sum_{k=0}^{\infty} \frac{(-1)^{2k+1}}{(2k+1)^2} (-1)^k
\]

- \( (-1)^{2k+1} = (-1)(-1)^{2k} = (-1) \), since \( (-1)^{2k} = 1 \).
- So, this simplifies to:
  \[
  \sum_{k=0}^{\infty} \frac{-1}{(2k+1)^2} (-1)^k = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(2k+1)^2}
  \]
  This is a convergent alternating series, often associated with a series expansion for sine functions at odd multiples of \( \frac{\pi}{2} \).

### Expression 2:
\[
\sum_{k=0}^{\infty} \frac{(-1)^{2k+1}}{(2k+1)^2} (-1)^{k+1}
\]

- Using the same simplification for \( (-1)^{2k+1} = -1 \), we get:
  \[
  \sum_{k=0}^{\infty} \frac{-1}{(2k+1)^2} (-1)^{k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}
  \]

### Comparison:

- **Expression 1** results in: \( \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(2k+1)^2} \), which has alternating signs starting with a negative term.
- **Expression 2** results in: \( \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2} \), which alternates but starts with a positive term.

For the sine function at odd multiples of \( \frac{\pi}{2} \), we typically expect an alternating series that begins with a positive term, so **Expression 2** is the correct representation.

### Conclusion:
The correct expression is:
\[
\sum_{k=0}^{\infty} \frac{(-1)^{2k+1}}{(2k+1)^2} (-1)^{k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}
\]

Do you want more details or have any further questions?

Here are 5 related questions to expand on this concept:
1. How does this series relate to the Fourier series for \( \sin(x) \)?
2. What is the convergence behavior of alternating series in general?
3. Can this series be related to any integral representation of \( \sin\left(\frac{\pi}{2}\right) \)?
4. How would the series change if it were for \( \cos(x) \) instead of \( \sin(x) \)?
5. How is this series connected to the Riemann zeta function for specific values?

**Tip**: Always simplify powers of \( (-1) \) early in the problem to reduce the complexity of the expression.

Which is correct, representing sin(\frac{(2k+1)pi}{2}) as \sum_{k=0}^{\infty} \frac{(-1)^{2k+1}}{(2k+1)^2} (-1)^k or \sum_{k=0}^{\infty} \frac{(-1)^{2k+1}}{(2k+1)^2} (-1)^{k+1}?

Discover the correct series expansion of sin(\frac{(2k+1)\pi}{2}) and explore the alternating series that correctly represents this function. This explanation covers trigonometric series expansions, alternating series, and Fourier series for sine functions. Suitable for undergraduate students studying calculus or higher-level mathematics.

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