91Ó°ÊÓ

Understanding the convergence of series is crucial in calculus. A series is an infinite sum of terms given by a sequence. We mainly focus on whether this sum reaches a finite limit—a property known as convergence. To test convergence, we often use the Comparison Test, which allows us to compare a series to another known convergent series. If the terms of our series are smaller than the terms of a known convergent series, the original series also converges. This technique is beneficial because it lets us simplify complicated series by relating them to easier or well-known ones.

In our problem, the first series \(\sum_{k=1}^{\infty} \frac{1}{3^k + 5}\) is compared with the series \(\sum_{k=1}^{\infty} \frac{1}{3^k}\), a geometric series. Similarly, the second series \(\sum_{k=1}^{\infty} \frac{5 \sin^2 k}{k !}\) is compared with \(\sum_{k=1}^{\infty} \frac{5}{k!}\), related to the exponential series. This method helps confirm that both original series are convergent.

A geometric series is one of the simplest forms of series that we can work with easily in calculus. It has a specific pattern where each term is a constant multiple of the previous term. For a geometric series: \(\sum_{k=1}^{\infty} ar^k\), we have:

• \(a\) is the first term,
• \(r\) is the common ratio.

The series converges if the absolute value of the ratio \(|r| < 1\). The sum of an infinite convergent geometric series can be calculated using the formula: \(\frac{a}{1-r}\).

In our exercise, the series \(\sum_{k=1}^{\infty} \frac{1}{3^k}\) is a simple geometric series with \(a = 1\) and \(r = \frac{1}{3}\), ensuring convergence since \(\frac{1}{3} < 1\). By comparing another series to this convergent geometric series, we can determine the original series also converges.

Factorials appear frequently in series, especially in problems involving combinations and permutations. A factorial, noted as \(n!\), is the product of all positive integers from 1 up to \(n\). Factorials grow extremely fast as \(n\) increases, making them handy when evaluating series for convergence.

For instance, the series \(\sum_{k=1}^{\infty} \frac{1}{k!}\) closely resembles the exponential series \(e^x\), evaluated at \(x = 1\). This makes it convergent due to the rapid growth of \(k!\), which ensures that series with terms \(\frac{1}{k!}\) decrease rapidly. In the exercise, the series \(\sum_{k=1}^{\infty} \frac{5 \,}{k!}\) is shown to converge due to the nature of factorials in the denominator, and serves as a comparison to assess the convergence of another similar series.