Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \sum_{n=1}^{\infty} \frac{1}{n \sqrt{n+1}} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt{n+1}} \) converges.

Step 1: Understand the Series

The given series is \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt{n+1}} \). This is an infinite series where each term is of the form \( \frac{1}{n \sqrt{n+1}} \). The denominator grows as \( n \) increases, indicating that the terms become smaller as \( n \) goes to infinity.

Step 2: Determine the Nature of the Series

We need to determine if this series converges or diverges. To do this, we often use convergence tests suitable for infinite series with positive terms like the Comparison Test, Limit Comparison Test, or Integral Test.

Step 3: Use the Limit Comparison Test

For the Limit Comparison Test, we compare our series with a simpler series. Let's compare \( \frac{1}{n \sqrt{n+1}} \) with \( \frac{1}{n^{3/2}} \). This is because as \( n \to \infty \), \( \sqrt{n+1} \approx \sqrt{n} \), so \( \frac{1}{n \sqrt{n+1}} \approx \frac{1}{n^{3/2}} \).

Step 4: Perform the Limit Comparison Test

Calculate \( \lim_{n \to \infty} \frac{\frac{1}{n \sqrt{n+1}}}{\frac{1}{n^{3/2}}}\). This limit simplifies to \( \lim_{n \to \infty} \frac{n^{3/2}}{n \sqrt{n+1}} = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n}}} \). As \( n \to \infty \), the expression inside the square root approaches 1, so the limit is 1.

Step 5: Compare with the P-Series

Since the limit is a finite non-zero number (1), by the Limit Comparison Test, the given series \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt{n+1}} \) has the same convergence behavior as \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \). The series \( \sum_{n=1}^{\infty} \frac{1}{n^{p}} \) converges when \( p > 1 \). Here, \( p = \frac{3}{2} > 1 \), so \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) converges.

Step 6: Conclusion

Using the Limit Comparison Test, we have shown that the original series \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt{n+1}} \) converges because it is similar to the convergent p-series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \). Thus, the series converges.

Key Concepts

Convergence Tests

Convergence tests are methods used to determine if an infinite series converges or diverges. When we say a series converges, we mean that as you add up more and more terms in the series, the sum approaches a specific finite number. On the other hand, a diverging series does not settle to any particular value.
Some common convergence tests include:

• The Comparison Test, which involves comparing the terms of your series to another series with known convergence behavior.
• The Limit Comparison Test, where we take the limit of the ratio of the terms from two series as the index goes to infinity.
• The Integral Test, which compares the series to an integral to determine convergence.

Selecting the right convergence test can simplify the process of determining if a series converges. In particular, it is often useful to choose a test based on how the series behaves for large values of the index.

Limit Comparison Test

The Limit Comparison Test is useful when you can find another series with similar behavior to the one you are studying. To use this test, you choose a series that you already know converges or diverges. Then, you compute the limit of the ratio of the general terms of your series and the comparison series.
Here's how it works:

• Let the original series be \( \sum_{n=1}^{\infty} a_n \) and the comparison series be \( \sum_{n=1}^{\infty} b_n \).
• Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
• If this limit is a positive finite number, then both series either converge or diverge together.

In our example, we compared \( \frac{1}{n \sqrt{n+1}} \) to the simpler series \( \frac{1}{n^{3/2}} \) and found that the limit is 1, indicating that both series have the same convergence behavior.

P-Series

P-series are a special type of series which have the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Understanding p-series is vital, as they serve as benchmarks for comparison in tests like the Limit Comparison Test.
The convergence of a p-series depends on the value of \( p \):

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

In our example, we found a p-series with \( p = \frac{3}{2} \), which converges since the exponent \( \frac{3}{2} > 1 \). Thus, by comparing to this p-series, we have shown that our original series also converges.

Mathematical Analysis

Mathematical analysis forms the foundation upon which convergence tests for infinite series are built. It involves studying the limits, functions, derivatives, and many other concepts that help us understand the behavior of mathematical expressions and structures.
In essence:

• Analysis provides tools to rigorously determine how expressions behave as variables approach certain limits.
• It allows us to formalize concepts like convergence, continuity, and differentiability.
• For series, analysis helps establish when a sequence of partial sums converges to a specific value.

The exercise on infinite series is a classic application of mathematical analysis, as it requires us to rigorously justify the convergence of terms using methodical approaches like the Limit Comparison Test and understanding series types such as p-series.