Question

Use the comparison test to determine whether the following series converge. $$ \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n} $$

Short answer

The series converges by the comparison test.

Step-by-step solution

Understand the series

The series under consideration is \( \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n} \). We want to determine whether this series converges or diverges using the comparison test.

Simplify the term

To simplify \( \sqrt{n+1}-\sqrt{n} \), multiply and divide by the conjugate \( \sqrt{n+1} + \sqrt{n} \). This gives:\[\sqrt{n+1}-\sqrt{n} = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}.\]

Substitute simplified term into the series

Substitute \( \sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} \) back into the original series. The series becomes:\[\sum_{n=1}^{\infty} \frac{1}{n(\sqrt{n+1} + \sqrt{n})}.\]

Compare with a known convergent series

Notice that \( \sqrt{n+1} + \sqrt{n} > \sqrt{n} + \sqrt{n} = 2\sqrt{n} \), thus \( \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} \). Therefore, each term in the series:\[\frac{1}{n(\sqrt{n+1} + \sqrt{n})} < \frac{1}{2n\sqrt{n}}.\]This resembles the series \( \sum \frac{1}{n^{3/2}} \), which is known to converge.

Apply the comparison test

The comparison test states that if \( 0 \leq a_n \leq b_n \) for all relevant \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) converges as well. Since \( \frac{1}{2n\sqrt{n}} \) converges and \( \frac{1}{n(\sqrt{n+1} + \sqrt{n})} < \frac{1}{2n\sqrt{n}} \), by the comparison test, the series \( \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n} \) also converges.

Key concepts

Series Convergence

When dealing with infinite series, we often need to determine if a series converges or diverges. Convergence means that as you continue to add terms from the series, it approaches some finite limit.
This is important because a converging series can often be used to represent some real quantities or values in various mathematical and scientific contexts.

• Convergence: A series \( \sum_{n=1}^{\infty} a_n \) is said to converge if the sequence of partial sums \( S_N = a_1 + a_2 + \cdots + a_N \) approaches a finite limit as \( N \to \infty \).
• Divergence: If the partial sums do not approach a finite limit, the series is said to diverge.

In our specific exercise, the series is expressed as \( \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n} \). Our task is to determine whether this series converges using the comparison test.

Algebraic Manipulation

Algebraic manipulation plays a crucial role in simplifying complex expressions within a sequence or series. In our problem, we needed to simplify the expression \( \sqrt{n+1} - \sqrt{n} \). This required multiplying and dividing by the conjugate \( \sqrt{n+1} + \sqrt{n} \). This technique simplifies our initial complex difference of square roots into a more manageable expression.

• Original term: \( \sqrt{n+1} - \sqrt{n} \)
• Multiply and divide by the conjugate: \( \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}} \)
• Simplified term: \( \frac{1}{\sqrt{n+1} + \sqrt{n}} \)

Algebraic manipulation helps us rewrite expressions in ways that are easier to analyze, often turning a complex, abstract problem into something accessible and solvable.

Mathematical Inequalities

Mathematical inequalities help us compare sizes of expressions, which is fundamental in the application of the comparison test. In our exercise, we need to establish inequalities to relate our given series to a known convergent series.

Here's how it works:

• We found that \( \sqrt{n+1} + \sqrt{n} > 2\sqrt{n} \).
• This leads to \( \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} \).
• Therefore, the terms of our series \( \frac{1}{n(\sqrt{n+1} + \sqrt{n})} < \frac{1}{2n\sqrt{n}} \).

By showcasing this inequality, we can compare our series with a simpler series, \( \sum \frac{1}{n^{3/2}} \), which is known to converge. Thus, using the properties of inequalities and the comparison test, we confirm that our original series also converges.