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Inequalities are mathematical expressions that show the relationship between two values, indicating that one is larger or smaller than the other. Solving inequalities often involves finding the range of values that satisfy the inequality condition. In this exercise, we are tasked with finding a positive integer \( M \) such that for all \( n \geq M \), the inequality \( n^{10} \cdot 0.99^n < 0.995^n \) holds. To tackle this, identify where the left-hand side is less than the right-hand side. Initially, it may not be obvious. However, by analyzing how the terms behave as \( n \) increases, it becomes clearer that \( 0.99^n \) decreases much faster than \( n^{10} \) grows. This is the key observation that guides us in finding an \( M \). A reliable method to solve such inequalities is by reformulating them using logarithms or through suitable approximations. For example, transformation of the given inequality into logarithmic terms helps in estimating the value of \( M \) accurately. You seek a point where the power of \( n \) in the polynomial \( n^{10} \) becomes negligible compared to the exponential decay of \( 0.99^n \). This highlights the essential skill of comparing rates of growth in inequalities.

A geometric series is a sequence of numbers where each term is a fixed multiple (called the common ratio) of the previous term. The formula for the sum of an infinite geometric series is given as \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. For the series to converge, the common ratio \( r \) must satisfy \(|r| < 1\). In our exercise, we compare the series \( \sum_{n=1}^{\infty} n^{10} \cdot 0.99^n \) to a geometric series \( \sum_{n=1}^{\infty} 0.995^n \), which is known to converge due to its common ratio \( 0.995 < 1 \). Geometric series are straightforward to understand because the ratio between successive terms remains constant. This property makes it easy to assess convergence. If a more complicated series is bounded by a geometric series with a common ratio less than 1, as is done in our exercise, it also converges. Understanding geometric series and their convergence is pivotal because they provide a baseline against which more complex series can be compared, utilizing their well-known behavior to make conclusions about other series.

The Comparison Test is a method used to determine the convergence or divergence of an infinite series by comparing it to another series with known behavior. This test is powerful because it leverages our understanding of simpler series, like geometric or p-series, to draw conclusions about more complicated series.In this problem, we apply the Comparison Test by comparing \( \sum_{n=1}^{\infty} n^{10} \cdot 0.99^n \) with the geometric series \( \sum_{n=1}^{\infty} 0.995^n \). We already know that \( 0.995^n \) decreases faster than \( n^{10} \cdot 0.99^n \) grows for sufficiently large \( n \) (specifically for \( n \geq M \)). Since the geometric series \( \sum 0.995^n \) converges, the Comparison Test tells us that our given series must also converge because its terms become smaller than those of the convergent series.The ability to use the Comparison Test effectively requires a strong grasp on the rate at which different types of functions approach infinity or zero. Recognizing when a series can be compared successfully, and correctly setting the bounds, is essential for using this test. It is a handy tool in calculus for analyzing the convergence of series beyond just those with geometric or arithmetic terms.