The given summation is:

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$

This series alternates in sign due to the factor $(-1)^n$ and involves the reciprocals of the squares of integers. Let's analyze this in steps:

Step 1: Comparison to a known series

The series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is known as the Basel problem, and its sum is:

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

Now, in the given summation, the term alternates in sign, so we have to take that into account.

Step 2: Alternating Series Representation

The given series:

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$

is an alternating series of the form $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$.

This alternating version of the Basel problem can be computed through known results in mathematical analysis. It turns out that:

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} = -\frac{\pi^2}{12}$

Step 3: Explanation for the Result

The alternating series introduces a factor that reduces the sum of the series to half the value of the Basel problem but with a negative sign. This happens because alternating the signs introduces cancellation between consecutive terms, leading to the halving effect.

Final Answer:

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} = -\frac{\pi^2}{12}$

Would you like more details or have any questions about this solution?

Here are 5 related questions you might find interesting:

1. What is the general form of an alternating series, and why does it converge?
2. How is the Basel problem solved, and what techniques are used in its derivation?
3. How can you approximate $\pi$ using series like the one we just solved?
4. What other famous series involve $\pi$ in their sums?
5. What are the conditions for the convergence of an alternating series?

Tip: Alternating series often simplify results by canceling terms, making them useful in approximations.