91Ó°ÊÓ

To prove that a sequence converges, we use the limit definition of convergence. This means we need to show that the terms of the sequence get arbitrarily close to a specific value (called the limit) as the sequence goes on forever. For a sequence \((a_n)\) to converge to a limit \(L\), for every positive number \(\epsilon\), no matter how small, there exists a number \(N\) such that for all \(n > N\), the inequality \(|a_n - L| < \epsilon\) holds. This tells us that beyond a certain point in the sequence, all the terms are very close to \(L\). Applying this to our specific sequence \(a_n = n r^n\), we aim to show that for any \(\epsilon > 0\), we can find an \(N\) where for all \(n > N\), \(|n r^n - 0| < \epsilon\). Simplifying, this becomes \(|n r^n| < \epsilon\). Understanding this definition is crucial for proving convergence of any sequence.

A geometric sequence has the form \(a, ar, ar^2, ar^3, \ldots\) Each term is obtained by multiplying the previous term by a fixed number \(r\), called the common ratio. In our sequence \(a_n = n r^n\), the term \(r^n\) plays a critical role. When \(|r| < 1\), the powers of \(r\) get smaller and smaller as \(n\) increases. This occurs because a fraction multiplied by itself repeatedly results in an even smaller fraction. Consider an example where \( r = \frac{1}{2}\). As \(n\) grows, \( (\frac{1}{2})^n \) approaches zero very quickly: \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \cdots \). This behavior of \( r^n\) approaching zero exponentially fast is a key observation in understanding why \( n r^n \) converges to zero despite the presence of the increasing term \(n\).

The epsilon-delta definition is a precise method used to show that a sequence converges to a limit. It states that for any \( epsilon > 0\), there must exist a corresponding \( N \) such that for all terms \(n > N\), the absolute difference \(| a_n - L | < \epsilon\). In the context of our sequence, \( a_n = n r^n \), we need to find an \( N \) where \(n r^n < \epsilon\). Given that \(| r | < 1\), \( r^n \) becomes very small for large \( n \), allowing us to control the growth of \(n\). We can choose a large enough \(N\) such that \(n > N\) ensures \(r^n < \frac{\epsilon}{n}\). For large \(n\), \(r^n\) decreases faster than \(n\) increases, making \(n r^n \) ultimately less than \( epsilon\), proving the convergence. This critical step follows directly from understanding and utilizing the epsilon-delta definition.