Convergence Analysis Of Series With Arctan, Alternating Signs, And Sine Terms

Hey everyone! Today, we're going to dissect a fascinating problem from the realm of real analysis, specifically dealing with the convergence of series. We'll be tackling the series:

$$\sum_{n=1}^\infty \Bigl(\tfrac{\arctan n}{n} + (-1)^n\Bigr)\,\sin\!\Bigl(\tfrac1n\Bigr)$$

This series looks a bit intimidating at first glance, mixing trigonometric functions, alternating signs, and rational expressions. But don't worry, we'll break it down step by step and explore its absolute and conditional convergence. So, let's buckle up and dive in!

Understanding the Series Components

Before we jump into the convergence tests, let's get a good handle on the individual components of our series. This will give us a better intuition for how the series behaves as n grows larger.

The Arctangent Term: arctan(n)/n

The arctan function, or inverse tangent, is a key player here. Remember that arctan(x) gives you the angle whose tangent is x. As n approaches infinity, arctan(n) approaches π/2. This is because the tangent function approaches infinity as the angle approaches π/2. So, arctan(n) is bounded above by π/2.

Now, let's consider the term arctan(n)/n. We have a bounded function (arctan(n)) divided by n, which grows without bound. This means that arctan(n)/n approaches 0 as n tends to infinity. Think of it like this: you're dividing a number that's getting closer and closer to π/2 by an increasingly large number. The result will inevitably shrink towards zero. This behavior is crucial for our convergence analysis.

The Alternating Term: (-1)^n

The term (-1)^n introduces an alternating sign into the series. When n is even, (-1)^n is 1, and when n is odd, (-1)^n is -1. This creates an oscillating pattern in the series, where terms alternate between positive and negative values. Alternating series often exhibit conditional convergence, which we'll explore later. This oscillation can sometimes help a series converge, even if the absolute values of the terms don't converge to zero quickly enough.

The Sine Term: sin(1/n)

Finally, we have sin(1/n). As n approaches infinity, 1/n approaches 0. We know that sin(x) behaves very similarly to x for small values of x. More formally, we can say that lim (x→0) sin(x)/x = 1. This is a fundamental limit in calculus. Therefore, sin(1/n) behaves like 1/n as n gets large. This is a crucial observation because it allows us to compare our series with simpler series whose convergence properties we already know, like the harmonic series.

Understanding this asymptotic behavior is key to tackling the problem. The sine function provides a link to the well-understood behavior of the harmonic series, which is a cornerstone in convergence analysis.

Breaking Down the Series

Now that we've analyzed the individual components, let's rewrite the series to make it easier to work with. We can distribute the sin(1/n) term across the parentheses:

$$\sum_{n=1}^\infty \Bigl(\tfrac{\arctan n}{n} + (-1)^n\Bigr)\,\sin\!\Bigl(\tfrac1n\Bigr) = \sum_{n=1}^\infty \frac{\arctan n}{n} \sin\!\Bigl(\tfrac1n\Bigr) + \sum_{n=1}^\infty (-1)^n \sin\!\Bigl(\tfrac1n\Bigr)$$

This splits our original series into two separate series. Let's call them Series I and Series II:

• Series I: $\sum_{n=1}^\infty \frac{\arctan n}{n} \sin\!\Bigl(\tfrac1n\Bigr)$
• Series II: $\sum_{n=1}^\infty (-1)^n \sin\!\Bigl(\tfrac1n\Bigr)$

By analyzing the convergence of each of these series separately, we can determine the convergence behavior of the original series. If both series converge, then the original series converges. If one converges and the other diverges, then the original series diverges. If both diverge, we need to investigate further to see if they diverge in a way that cancels each other out, leading to convergence (which is less common but possible).

Analyzing Series I: Absolute Convergence

Let's tackle Series I first: $\sum_{n=1}^\infty \frac{\arctan n}{n} \sin\!\Bigl(\tfrac1n\Bigr)$. To determine its convergence, we'll use the Comparison Test. This test involves comparing our series to a known convergent or divergent series.

Using the Comparison Test

Remember that arctan(n) is bounded above by π/2 and sin(1/n) behaves like 1/n for large n. This gives us the following inequality:

$$\frac{\arctan n}{n} \sin\!\Bigl(\tfrac1n\Bigr) \leq \frac{\pi/2}{n} \cdot \frac{1}{n} = \frac{\pi}{2n^2}$$

So, we've bounded the terms of Series I above by π/(2n^2). Now, let's consider the series $\sum_{n=1}^\infty \frac{\pi}{2n^2}$. This is a constant multiple of the p-series $\sum_{n=1}^\infty \frac{1}{n^2}$, where p = 2.

The P-Series Test

The p-series test is a powerful tool for determining the convergence of series of the form $\sum_{n=1}^\infty \frac{1}{n^p}$. The test states that the series converges if p > 1 and diverges if p ≤ 1. In our case, p = 2, which is greater than 1. Therefore, the series $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, and so does $\sum_{n=1}^\infty \frac{\pi}{2n^2}$ (since multiplying a convergent series by a constant doesn't change its convergence).

By the Comparison Test, since Series I is bounded above by a convergent series, Series I converges absolutely. Absolute convergence is a strong type of convergence; it means that not only does the series converge, but the series formed by taking the absolute values of its terms also converges. This is a key piece of information for understanding the overall behavior of our original series. Guys, this was a great use of the Comparison Test!

Analyzing Series II: Conditional Convergence

Now, let's turn our attention to Series II: $\sum_{n=1}^\infty (-1)^n \sin\!\Bigl(\tfrac1n\Bigr)$. This is an alternating series, which suggests we should consider the Alternating Series Test.

The Alternating Series Test

The Alternating Series Test provides conditions for the convergence of a series of the form $\sum_{n=1}^\infty (-1)^n b_n$ (or $\sum_{n=1}^\infty (-1)^{n+1} b_n$), where b_n are positive terms. The test has two main requirements:

1. b_n must be a decreasing sequence.
2. lim (n→∞) b_n = 0.

If both of these conditions are met, then the alternating series converges.

In our case, b_n = sin(1/n). Let's check if these conditions are satisfied.

Checking the Conditions

1. Decreasing Sequence: We need to show that sin(1/n) is a decreasing sequence. As n increases, 1/n decreases. Since the sine function is increasing on the interval [0, π/2], and 1/n is always in this interval for n ≥ 1, it follows that sin(1/n) is a decreasing sequence.
2. Limit to Zero: We need to show that lim (n→∞) sin(1/n) = 0. As n approaches infinity, 1/n approaches 0. Since sin(0) = 0, we have lim (n→∞) sin(1/n) = 0.

Both conditions of the Alternating Series Test are satisfied! Therefore, Series II converges. However, this convergence is conditional, meaning that the series converges, but the series formed by taking the absolute values of its terms does not converge.

Conditional vs. Absolute Convergence

To understand why Series II is only conditionally convergent, let's consider the series of absolute values: $\sum_{n=1}^\infty |(-1)^n \sin\!\Bigl(\tfrac1n\Bigr)| = \sum_{n=1}^\infty \sin\!\Bigl(\tfrac1n\Bigr)$. As we discussed earlier, sin(1/n) behaves like 1/n for large n. The series $\sum_{n=1}^\infty \frac{1}{n}$ is the harmonic series, which is a classic example of a divergent series. Therefore, the series of absolute values diverges, and Series II is only conditionally convergent. Understanding this distinction between conditional and absolute convergence is super important!

Final Verdict: Convergence of the Original Series

We've analyzed both Series I and Series II. We found that:

• Series I converges absolutely.
• Series II converges conditionally.

Since Series I converges and Series II converges, their sum also converges. Therefore, the original series,

$$\sum_{n=1}^\infty \Bigl(\tfrac{\arctan n}{n} + (-1)^n\Bigr)\,\sin\!\Bigl(\tfrac1n\Bigr)$$

converges conditionally. It converges because the alternating term helps to tame the divergence that might otherwise occur due to the 1/n behavior of the sine function. However, it doesn't converge absolutely because the positive terms, behaving like the harmonic series, add up too slowly to reach a finite sum when considered alone. This was quite the journey, but we made it! Understanding the interplay between the arctangent, alternating sign, and sine terms was key to unlocking the solution.

Repair Input Keyword

Examine the absolute and conditional convergence of the series $\sum_{n=1}^\infty \Bigl(\tfrac{\arctan n}{n} + (-1)^n\Bigr)\,\sin\!\Bigl(\tfrac1n\Bigr)$.

Title

Convergence Analysis of Series with Arctan, Alternating Signs, and Sine Terms