Problem 10 - Analysis with an introduction to proof

Question

Determine whether or not the series \(\sum_{n=1}^{\infty} 1 /(\sqrt{n+1}+\sqrt{n})\) converges. Justify your answer. ??

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1} + \sqrt{n}}\) diverges because it satisfies the inequality \(\frac{1}{\sqrt{n+1} + \sqrt{n}} \geq \frac{1}{n}\) and, using the comparison test with the well-known divergent harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), it is shown to also diverge.

Step 1: Rewrite the given series

First, let's rewrite the given series in a more convenient form: \[\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1} + \sqrt{n}}\]

Step 2: Compare to another series

Now, let's compare it to the series \(\sum_{n=1}^{\infty} \frac{1}{n}\) as it is a well-known fact that harmonic series diverges. In order to compare the series, let's show that: \[\frac{1}{\sqrt{n+1} + \sqrt{n}} \geq \frac{1}{n}\]

Step 3: Simplify the inequality

We can simplify the inequality by cross-multiplying and rearranging the terms: \[n(\sqrt{n+1} + \sqrt{n}) \geq (\sqrt{n+1} + \sqrt{n})\] Since \(\sqrt{n+1} + \sqrt{n} > 0\), we can safely divide both sides of the inequality by \((\sqrt{n+1} + \sqrt{n})\) without reversing the inequality sign \[n \geq 1\]

Step 4: Apply the comparison test

As shown, every term of our original series is greater than or equal to the corresponding term of the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which is a divergent series. Using the comparison test, because the harmonic series diverges and every term in the original series is greater or equal to the corresponding term in the harmonic series, the original series also diverges.

Step 5: Conclusion

Thus, we conclude that the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1} + \sqrt{n}}\) diverges.

Key Concepts

Divergent Series

When we talk about divergent series, we refer to a type of mathematical series that does not converge to a limit. In simpler terms, the sum of a divergent series keeps increasing or decreasing indefinitely as we keep adding more terms. This is in contrast to a convergent series, where the sum approaches a specific value.

• A series is said to diverge if its terms do not sum to a finite number.
• Divergence can be identified using various tests, such as the Comparison Test.
• The harmonic series is a classic example of a divergent series.

In the given exercise, we determine if a series diverges by comparing it to the harmonic series, which is already known to diverge. By doing this, we can establish that if our series is greater than or equal to another known divergent series at each term, it too must diverge.

Comparison Test

The Comparison Test is a useful tool in determining the convergence or divergence of a series. It involves comparing a series of interest to another series with known behavior.
Here's how it works in simple terms:

• If a series with positive terms is less than a convergent series, then it is also convergent.
• If a series with positive terms is greater than or equal to a divergent series, then it diverges.

To apply the test, you must find a known series and show that every term of your series is consistently larger or smaller than the terms of the known one.
In our example, we compared our series to the harmonic series. By showing that each term in the original series is greater or equal to the corresponding term in the harmonic series, we concluded the original series diverges using the Comparison Test.

Harmonic Series

The harmonic series is essential in understanding many series' convergence or divergence. It is defined as follows:\[ ext{Harmonic Series: } \, \ \sum_{n=1}^{\infty} \frac{1}{n} \]This series is a classic example of a divergent series because its sum grows without bound, even though its terms become smaller.
A few characteristics of the harmonic series are:

• It is an infinite series of reciprocals of natural numbers.
• Despite its terms approaching zero, the sum of the series never levels out to a finite limit.
• Its divergence is often proven using comparison or integral tests.

In our problem, the harmonic series served as the benchmark for determining divergence. By comparison, if every term in another series is more significant or equal to the harmonic series, the series in question will diverge.