(a) Using computer or tables (or see Chapter 7 , Section 11 ), verify that \(\sum_{n=1}^{\infty} 1 / n^{2}=\) \(\pi^{2} / 6=1.6449+,\) and also verify that the error in approximating the sum of the series by the first five terms is approximately 0.1813 . (b) \(\quad\) By computer or tables verify that \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)(1 / 2)^{n}=\pi^{2} / 12-(1 / 2)(\ln 2)^{2}=\) \(0.5822+,\) and that the sum of the first five terms is \(0.5815+.\) (c) Prove theorem (14.4). Hint: The error is \(\left|\sum_{N+1}^{\infty} a_{n} x^{n}\right| .\) Use the fact that the absolute value of a sum is less than or equal to the sum of the absolute values. Then use the fact that \(\left|a_{n+1}\right| \leq\left|a_{n}\right|\) to replace all \(a_{n}\) by \(a_{N+1},\) and write the appropriate inequality. Sum the geometric series to get the result.

In mathematics, an infinite series is a sum of an infinite sequence of terms. The series can be represented as \(\textstyle \sum_{n=1}^{\infty} a_n\). To understand the behavior of an infinite series, we need to know whether it converges (approaches a specific value) or diverges (does not approach any limit).
When a series converges, we can capture the total value by adding up all its terms, no matter how many there are. For example, the series \(\textstyle \sum_{n=1}^{\infty} \frac{1}{n^2}\) sums up to \[\frac{\pi^2}{6} \approx 1.6449\].

To approximate an infinite series, we sum a large number of its terms. However, there's always some error in finite approximations. In the problem given, summing the first five terms of \(\textstyle \sum_{n=1}^{\infty} \frac{1}{n^2}\) gives 1.4636, and the error can be found by subtracting this sum from the actual infinite sum.

Series approximation involves estimating the sum of an infinite series by summing a finite number of terms. This technique is crucial when dealing with series that cannot be easily summed to infinity.
Sometimes, the error of such an approximation can be calculated or estimated. For the series \(\textstyle \sum_{n=1}^{\infty} \frac{1}{n^2} \left( \frac{1}{2}\right)^n \), the first five terms give a sum of approximately 0.5815, which closely approximates 0.5822, indicating the approximation's accuracy.

Errors in series approximations arise because we don’t include every term in the series. The remaining part is known as the remainder or the error term. For instance, the error term for the series \(\textstyle \sum_{n=1}^{5} \frac{1}{n^2} \left(\frac{1}{2}\right)^n \) is the difference between the actual infinite series sum (0.5822) and the approximation (0.5815). To minimize this error, more terms need to be summed.

A geometric series is a series with a constant ratio between successive terms. It is given by the formula \(\textstyle a + ar + ar^2 + ar^3 + \ldots\), where \(a\) is the first term and \(r\) is the common ratio.
If \( |r| < 1\), the series converges to \[ \frac{a}{1-r} \].

Considering the geometric nature of the series in the exercise, we can notice that when \(r = 0.5\), the series converges. This property is employed when summing partial series, such as \(\textstyle \sum_{n=1}^{5} \frac{1}{n^2} \left(\frac{1}{2}\right)^n \). Each term decreases rapidly, making partial sums quite accurate even with just a few terms added.
Recognizing series as geometric helps in using convergence criteria and summing formulas for quick calculations.