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.