Problem 26 - Single Variable Calculus

Question

Determine whether the series is convergent or divergent. \(\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^{5}+n}}\)

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^{5}+n}}\) is convergent, as determined by the Comparison Test with a simpler convergent p-series, \(\sum_{n=1}^{\infty} n^{-\frac{3}{2}}\).

Step 1: Write down the given series

Let's write down the given series: \[ \sum_{n=1}^{\infty} \frac{n}{\sqrt{n^{5}+n}} \]

Step 2: Simplify the denominator

In order to compare the given series to a simpler series, first we need to simplify the expression. Let's factor the denominator: \[ \sqrt{n^5+n} = \sqrt{n(n^4+1)} \]

Step 3: Create a simpler series for comparison

Now we will create a simpler series that will be used for comparison. We will do this by removing the "+1" in the denominator as it has a minor effect on the series when n is large: \[ \sum_{n=1}^{\infty} \frac{n}{\sqrt{n(n^4+1)}} \approx \sum_{n=1}^{\infty} \frac{n}{\sqrt{n^5}} \]

Step 4: Simplify the approximate series further

Now, we will simplify the new series further: \[ \sum_{n=1}^{\infty} \frac{n}{\sqrt{n^5}} = \sum_{n=1}^{\infty} \frac{n}{n^{\frac{5}{2}}} = \sum_{n=1}^{\infty} n^{-\frac{3}{2}} \]

Step 5: Determine the convergence of the simpler series

The new series is now a p-series which converges if \(p > 1\) and diverges otherwise. In this case, \(p = \frac{3}{2}\) which is greater than 1, so the simpler series converges.

Step 6: Apply the Comparison Test

Now that we know the simpler series converges, we can apply the Comparison Test. The Comparison Test states that if \(0 \leq a_n \leq b_n\) for all n and \(\sum_{n=1}^{\infty} b_n\) converges, then \(\sum_{n=1}^{\infty} a_n\) also converges. Let's compare our original series (\(a_n\)) to the simpler series (\(b_n\)): \[ 0 \leq \frac{n}{\sqrt{n^5+n}} \leq \frac{n}{\sqrt{n^5}} = n^{-\frac{3}{2}} = b_n \] The inequality is true for all \(n >= 1\), and we know that the simpler series converges. Therefore, by the Comparison Test, the original series also converges.

Step 7: Conclusion

The series \(\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^{5}+n}}\) is convergent.

Key Concepts

Understanding the Comparison Test

The Comparison Test is a technique to determine if an infinite series converges or diverges. It works by comparing the series in question to another series whose convergence is already known. If the series you're examining is "smaller" than a known converging series, or "larger" than a known diverging series, you can infer the behavior of your original series.

• First, identify your two series: \(a_n\) (the one you're interested in) and \(b_n\) (the one you know about).
• If \(0 \leq a_n \leq b_n\) for all n and \sum b_n\ converges, then \sum a_n\ also converges.
• Conversely, if \a_n \geq b_n \ for all n and \sum b_n\ diverges, then \sum a_n\ diverges as well.

This method helps simplify the problem by comparing it to something already understood. It is essential when dealing with complex series.

Exploring p-series

A p-series is a specific type of infinite series given by the formula \sum_{n=1}^{\infty} \frac{1}{n^p}\. This series is handy because its convergence depends solely on the value of \(p\).

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

The p-series is fundamental when analyzing other complex series. In our example, the transformed series became \sum n^{-\frac{3}{2}}\, which has \(p = \frac{3}{2} > 1\) and thus converges. Recognizing when a series can transform into a p-series simplifies determining its convergence.

Steps for Simplification of Series

Simplifying a series is crucial when using the Comparison Test since it allows you to find a comparable series. Here's how you can simplify a series, as demonstrated in the provided solution:

• Identify the dominant terms in the expression. For example, in \sqrt{n^5+n}\, \(n^5\) dominates because it grows faster as \(n\) increases.
• Neglect less significant terms when \(n\) is large. Here, \(+1\) becomes negligible.
• Transform the expression to make comparisons easier, e.g., \sum \frac{n}{\sqrt{n^5}}\ simplifies to \sum n^{-\frac{3}{2}}\.

This method helps reveal underlying structures in series that are not immediately apparent. Simplification clears the path for applying tests effectively.

Understanding Infinite Series

Infinite series are sums of infinitely many terms. They often seem abstract but are foundational in mathematics. An infinite series is expressed as \sum_{n=1}^{\infty} a_n\, where \(a_n\) are terms of the series.

• Convergence means the series approaches a finite value as more terms are added.
• Divergence implies the series grows without bound or oscillates.

Infinite series appear in calculus, where they represent functions as sums, and in real-world applications such as computing areas under curves. Understanding whether a series converges or diverges helps in determining the behavior of functions represented by these series.