Problem 43 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n}(\sqrt{n+\sqrt{n}}-\sqrt{n}) $$

Step-by-Step Solution

Answer

The series converges conditionally but not absolutely.

Step 1: Simplify the Expression

First, let's simplify the term \(\sqrt{n+\sqrt{n}} - \sqrt{n}\) using the identity \((a-b)(a+b) = a^2 - b^2\). We can multiply and divide by the conjugate \(\sqrt{n+\sqrt{n}} + \sqrt{n}\):\[(\sqrt{n+\sqrt{n}} - \sqrt{n}) \cdot \frac{\sqrt{n+\sqrt{n}} + \sqrt{n}}{\sqrt{n+\sqrt{n}} + \sqrt{n}} = \frac{(n+\sqrt{n}) - n}{\sqrt{n+\sqrt{n}} + \sqrt{n}}\]This simplifies to:\[\frac{\sqrt{n}}{\sqrt{n+\sqrt{n}} + \sqrt{n}}\]

Step 2: Estimate the Simplified Expression

Now, \(\frac{\sqrt{n}}{\sqrt{n+\sqrt{n}} + \sqrt{n}}\) can be simplified further:- Notice that \(\sqrt{n+\sqrt{n}} \approx \sqrt{n}\). Therefore:\[\sqrt{n+\sqrt{n}} + \sqrt{n} \approx 2\sqrt{n}\]Thus, the expression becomes approximately:\[\frac{\sqrt{n}}{2\sqrt{n}} = \frac{1}{2}\]To refine it more, let's see the behavior as \(n\) becomes large. The equivalent expression behaves as small positive numbers as \(\frac{1}{2\cdot\sqrt{1 + \frac{1}{\sqrt{n}}}}\),which approximates to \(\frac{1}{2} (1 - \frac{1}{2} \cdot \frac{1}{\sqrt{n}})\).

Step 3: Consider Absolute Convergence

Check for absolute convergence by analyzing \(\sum_{n=1}^{\infty} \left| (-1)^{n} \left(\sqrt{n+\sqrt{n}} - \sqrt{n} \right) \right| \) which equals \(\sum_{n=1}^{\infty} \left(\sqrt{n+\sqrt{n}} - \sqrt{n} \right)\).Since each term approximates to \(\frac{1}{2\sqrt{n}}\), and \(\sum \frac{1}{\sqrt{n}}\) diverges (p-series with \(p=1/2\), which diverges), this new series also diverges.

Step 4: Test for Conditional Convergence

Since the series with absolute values diverges, let's use the Alternating Series Test on\[\sum_{n=1}^{\infty} (-1)^{n} \left(\sqrt{n+\sqrt{n}} - \sqrt{n} \right).\]The Alternating Series Test says a series \(\sum (-1)^n b_n\) converges if:1. \(b_n > 0\) for all \(n\), which ours is.2. \(b_n\) is decreasing, which can be seen, as each term tends to zero as \(n\) tends to infinity.3. \(\lim_{n \to \infty} b_n = 0\), already considered in previous steps.All conditions are satisfied, thus the series converges conditionally.

Key Concepts

Absolute Convergence

Absolute convergence is a concept that helps us understand whether a series converges regardless of the sign of its terms. To test for absolute convergence, we consider the series formed by the absolute values of the original series terms. If this new series converges, then the original series is said to converge absolutely. This is a strong form of convergence because it remains true even if the signs of the terms are changed or irregular.

In the given problem, we check for absolute convergence by observing \[\sum_{n=1}^{\infty} \left| (-1)^{n} \left(\sqrt{n+\sqrt{n}} - \sqrt{n} \right) \right| \]which essentially simplifies to checking the convergence of \[\sum_{n=1}^{\infty} \left(\sqrt{n+\sqrt{n}} - \sqrt{n} \right). \]
Both series involve positive terms. If this "absolute" series converges, the original series converges absolutely. However, in the exercise, since this absolute version approximates to a known divergent p-series \( \sum \frac{1}{\sqrt{n}} \), it also diverges. Thus, the given series does not converge absolutely.

Conditional Convergence

Conditional convergence occurs when a series converges, but its corresponding absolute series does not. This means that the convergence depends on the specific arrangement and alternating nature of the terms, which usually involve both positive and negative terms. This is more intricate than absolute convergence and relies on specific tests, such as the Alternating Series Test.

In our exercise, after discovering that the absolute series diverges, we explore the possibility of conditional convergence using the Alternating Series Test. The given series is \[\sum_{n=1}^{\infty} (-1)^{n} \left(\sqrt{n+\sqrt{n}} - \sqrt{n} \right) \]which presents an alternating form with terms going positive and negative. According to the Alternating Series Test, the series converges if:

• The sequence \(b_n = \sqrt{n+\sqrt{n}} - \sqrt{n} \) consists of positive terms, which it does.
• The terms are decreasing as \(n\) increases, which can be observed.
• The terms approach zero as \(n\) approaches infinity.

Since all these conditions are satisfied, the series converges conditionally.

p-Series Divergence

The p-series is a standard series in calculus defined as \[\sum_{n=1}^{\infty} \frac{1}{n^p} \]where \(p\) is a constant. The convergence of a p-series is determined by the exponent \(p\):

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

In the solution to the problem, the simplified series \(\sum \frac{1}{\sqrt{n}}\) aligns with a p-series where \(p = 0.5\). Since \(p = 0.5 < 1\), this series diverges. Understanding this concept allows us to recognize similar series configurations and quickly determine their convergence behavior. This step was crucial in the original problem to prove that the absolute series diverged, leading us to test for conditional convergence instead.