Question

In Exercises \(17-20\), approximate the sum of the series by using the first six terms. (See Example 4.) $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} 3}{n^{2}} $$

Short answer

The approximate sum of the series for the first six terms, keeping the structure and rules of an alternating series in mind, is calculated as follows: The first term, for \(n=1\), is 3. The second term, \(n=2\), is \(-\frac{3}{4}\). Continuing similarly for \(n=3\), \(n=4\), \(n=5\), and \(n=6\), one adds up all these terms to provide an approximate sum of the series.

Step-by-step solution

Understanding the terms of the series

It's important to first understand the terms of the series. We break down the series as follows: For \(n=1\), it will be \((-1)^{1+1} \times \frac{3}{1^2}\) which simplifies to \(\frac{3}{1}\). For \(n=2\), the segmentation results in \((-1)^{2+1} \times \frac{3}{2^2}\) which simplifies to \(-\frac{3}{4}\).We apply a similar process for the terms up to \(n=6\).

Summing up the first six terms

After understanding the series structure, the next step is to sum up the first six terms. This step involves calculating each term individually for \(n=1\) to \(n=6\) and then adding up these six term values.

Expressing the result

Bear in mind that this is an approximation of the sum of the series by considering only the first six terms. This sum is not the actual sum of the whole series, but just gives an approximate value.