$f(x) = \sum_{n=1}^{\infty} (-1)^n \frac{\sin \left( \frac{n \pi x}{2} \right)}{n^2}$ solve for when x=1 and this function is defined within the interval[0,2]

To evaluate the given series at $x = 1$ for the function:

\[
f(x) = \sum_{n=1}^{\infty} (-1)^n \frac{\sin \left( \frac{n \pi x}{2} \right)}{n^2}
\]

we can substitute $x = 1$ into the expression and analyze the sum.

### Step 1: Substituting $x = 1$ into the function
Substituting $x = 1$ in the expression:

\[
f(1) = \sum_{n=1}^{\infty} (-1)^n \frac{\sin \left( \frac{n \pi (1)}{2} \right)}{n^2}
= \sum_{n=1}^{\infty} (-1)^n \frac{\sin \left( \frac{n \pi}{2} \right)}{n^2}
\]

### Step 2: Simplify $\sin \left( \frac{n \pi}{2} \right)$
The values of $\sin \left( \frac{n \pi}{2} \right)$ repeat in a pattern as $n$ increases:

\[
\sin \left( \frac{n \pi}{2} \right) = 
\begin{cases}
0 & \text{for } n = 2, 4, 6, 8, \ldots \\
1 & \text{for } n = 1, 5, 9, 13, \ldots \\
-1 & \text{for } n = 3, 7, 11, 15, \ldots
\end{cases}
\]

Thus, we can rewrite the sum to reflect this periodic pattern of sine values.

### Step 3: Split the sum into relevant terms
The non-zero terms in the sum occur when $n = 1, 3, 5, 7, \dots$. Grouping the sine terms:

\[
f(1) = \sum_{k=0}^{\infty} (-1)^{2k+1} \frac{\sin \left( \frac{(2k+1) \pi}{2} \right)}{(2k+1)^2}
\]

For odd $n = 2k+1$, the sine function is either $1$ or $-1$, depending on $k$'s value.

### Step 4: Simplify the alternating series
Because $(-1)^{2k+1} = -1$ for all $k$, and the sine function alternates between $\pm1$, we now have:

\[
f(1) = - \left( \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \cdots \right)
\]

### Step 5: Recognize this as a known series
The series:

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

is a well-known series, and its sum is $\frac{\pi^2}{8}$. Thus:

\[
f(1) = - \frac{\pi^2}{8}
\]

### Final Answer:
\[
f(1) = - \frac{\pi^2}{8}
\]

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Would you like further details or have any questions?

Here are 5 related questions:
1. How is the sine function periodicity applied to infinite sums?
2. What are some other common alternating series, and how do we evaluate them?
3. How does the Leibniz series relate to this sum?
4. What is the role of Fourier series in representing such sums?
5. Can this function be represented as a Fourier series on a different interval?

**Tip:** Recognizing known series can save time in evaluating complex sums—many standard results are tabulated for quick reference!

Explore how to solve the infinite series $f(x) = \sum_{n=1}^{\infty} (-1)^n \frac{\sin(\frac{n \pi x}{2})}{n^2}$ for x=1. The solution involves trigonometric function periodicity and recognizing known series results. Suitable for Grades 11-12, this problem demonstrates the evaluation of alternating series and the application of sine function properties.

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