Convergence Test: Does The Series Converge Or Diverge?

Let's dive into the fascinating world of sequences and series to determine whether the given series converges or diverges. We are given the recursive definition of a sequence $a_n$ as follows:

$a_1 = 2$

$a_{n+1} = \frac{8 + \sin n}{n} a_n$

Our mission is to figure out the behavior of this sequence and, more importantly, whether the corresponding series converges or diverges. To do this, we'll explore the properties of the sequence, apply some convergence tests, and provide clear reasoning for our conclusion.

Analyzing the Sequence

To understand the behavior of the sequence, we need to look at the ratio $\frac{a_{n+1}}{a_n}$, which is given by:

$\frac{a_{n+1}}{a_n} = \frac{8 + \sin n}{n}$

Since $-1 \leq \sin n \leq 1$, we have $7 \leq 8 + \sin n \leq 9$. Therefore,

$\frac{7}{n} \leq \frac{8 + \sin n}{n} \leq \frac{9}{n}$

This inequality gives us some bounds on how the terms of the sequence change as $n$ increases. We can use this to analyze the convergence or divergence of the series.

Applying the Ratio Test

The ratio test is a powerful tool for determining the convergence or divergence of a series. It states that if we have a series $\sum a_n$, and we define $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$, then:

• If $L < 1$, the series converges.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

In our case, we have:

$\frac{a_{n+1}}{a_n} = \frac{8 + \sin n}{n}$

Let's find the limit as $n$ approaches infinity:

$L = \lim_{n \to \infty} \frac{8 + \sin n}{n}$

Since $-1 \leq \sin n \leq 1$, we know that $7 \leq 8 + \sin n \leq 9$. Thus, we can bound the expression:

$\lim_{n \to \infty} \frac{7}{n} \leq \lim_{n \to \infty} \frac{8 + \sin n}{n} \leq \lim_{n \to \infty} \frac{9}{n}$

Both $\lim_{n \to \infty} \frac{7}{n}$ and $\lim_{n \to \infty} \frac{9}{n}$ are equal to 0. Therefore, by the Squeeze Theorem,

$L = \lim_{n \to \infty} \frac{8 + \sin n}{n} = 0$

Since $L = 0 < 1$, the ratio test tells us that the series $\sum a_n$ converges. However, it is crucial to recognize what the ratio test actually implies. The ratio test determines the convergence or divergence of a series, not the sequence itself. The fact that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 0$ implies that the terms $a_n$ tend to zero faster than a geometric series with a ratio less than 1. This doesn't directly tell us about the convergence or divergence of the series. We need to investigate the behavior of the terms $a_n$ more carefully.

Analyzing the Terms $a_n$

Let's express $a_n$ in terms of $a_1$:

$a_2 = \frac{8 + \sin 1}{1} a_1$

$a_3 = \frac{8 + \sin 2}{2} a_2 = \frac{8 + \sin 2}{2} \cdot \frac{8 + \sin 1}{1} a_1$

In general,

$a_n = a_1 \prod_{k=1}^{n-1} \frac{8 + \sin k}{k} = 2 \prod_{k=1}^{n-1} \frac{8 + \sin k}{k}$

To determine whether the series $\sum a_n$ converges, we need to analyze the behavior of the product. We know that $7 \leq 8 + \sin k \leq 9$, so:

$a_n = 2 \prod_{k=1}^{n-1} \frac{8 + \sin k}{k} \leq 2 \prod_{k=1}^{n-1} \frac{9}{k} = 2 \cdot \frac{9^{n-1}}{(n-1)!}$

Now, let's consider the series $\sum_{n=1}^{\infty} \frac{9^{n-1}}{(n-1)!}$. This series is closely related to the exponential series. Recall that the Taylor series for $e^x$ is given by:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

So, if we let $x = 9$, we have:

$e^9 = \sum_{n=0}^{\infty} \frac{9^n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{9^n}{n!}$

Thus, the series $\sum_{n=1}^{\infty} \frac{9^{n-1}}{(n-1)!}$ converges because it's just a shifted version of the exponential series. Specifically, $\sum_{n=1}^{\infty} \frac{9^{n-1}}{(n-1)!} = e^9$.

Since $0 \leq a_n \leq 2 \cdot \frac{9^{n-1}}{(n-1)!}$ and $\sum_{n=1}^{\infty} \frac{9^{n-1}}{(n-1)!}$ converges, by the Comparison Test, the series $\sum_{n=1}^{\infty} a_n$ also converges.

Conclusion

The series defined by $a_1 = 2$ and $a_{n+1} = \frac{8 + \sin n}{n} a_n$ converges. We arrived at this conclusion by analyzing the terms of the sequence, bounding them using inequalities, and then applying the Comparison Test with a convergent exponential series.

In summary, understanding the behavior of recursive sequences requires a combination of analytical techniques and careful application of convergence tests. The interplay between the sequence definition and the properties of convergent series is crucial in determining the ultimate behavior of the series. By leveraging tools like the ratio test, comparison test, and knowledge of standard series like the exponential series, we can effectively analyze and determine the convergence or divergence of complex series.

To further enhance your understanding of series convergence and divergence, consider exploring resources such as those available on Khan Academy's Calculus section.