Question

Prove that the convergence of \(\Sigma a_{n}\) implies the convergence of $$\sum \frac{\sqrt{a_{n}}}{n}$$ if \(a_{n} \geq 0\)

Short answer

Answer: Yes, the convergence of \(\Sigma a_n\) implies the convergence of \(\sum \frac{\sqrt{a_n}}{n}\).

Step-by-step solution

State the given fact and define the series

It is given that the series \(\Sigma a_n\) converges, where \(a_n \geq 0\). The goal is to prove that the series $$\sum \frac{\sqrt{a_n}}{n}$$ also converges.

Apply the comparison test

To prove convergence of the series \(\sum \frac{\sqrt{a_n}}{n}\), the comparison test can be utilized. The comparison test states that if \(0 \leq b_n \leq a_n\) for all \(n\) and \(\sum a_n\) converges, then \(\sum b_n\) converges as well. First, notice that since \(a_n \geq 0\), it follows that \(\sqrt{a_n} \leq a_n\) for all \(n\), as the square root function is monotonically increasing for non-negative values. Therefore, the following inequality holds for all \(n\): $$0 \leq \frac{\sqrt{a_n}}{n} \leq \frac{a_n}{n}$$

Define a new series and apply the comparison test

Now define a new series, \(\sum b_n\), where \(b_n = \frac{a_n}{n}\). Notice that the inequality obtained in the previous step shows that \(0 \leq \frac{\sqrt{a_n}}{n} \leq b_n\) for all \(n\). To apply the comparison test, it is needed to show that \(\sum b_n\) converges. Using the integral test, it can be shown that the convergence of \(\sum a_n\) implies the convergence of \(\sum b_n\). The integral test states that if \(f(x)\) is a continuous, positive, and decreasing function on \([1,\infty)\), then the series \(\sum_{n=1}^\infty f(n)\) will converge if and only if the integral \(\int_1^\infty f(x) dx\) converges. Since \(a_n \geq 0\), the function \(f(x) = \frac{a_x}{x}\) is also always positive and decreasing, granted that \(a_n\) is a decreasing sequence. Therefore, the integral \(\int_1^\infty \frac{a_x}{x} dx\) converges. By the integral test, this implies that \(\sum b_n\) converges as well.

Conclude by comparison test

Since the series \(\sum b_n\) converges and the inequality \(0 \leq \frac{\sqrt{a_n}}{n} \leq b_n\) holds for all \(n\), the comparison test implies that the series \(\sum \frac{\sqrt{a_n}}{n}\) converges as well. In conclusion, the convergence of \(\Sigma a_n\) with \(a_n \geq 0\) implies the convergence of $$\sum \frac{\sqrt{a_n}}{n}$$

Key concepts

Comparison Test

The Comparison Test is a key method to determine the convergence of a series. It compares the series in question, let's say \(\sum b_n\), with another known series \(\sum a_n\). If \(0 \leq b_n \leq a_n\) for every term, and the series \(\sum a_n\) is known to converge, then \(\sum b_n\) must also converge. This is because if all terms of a smaller series \(b_n\) are dominated by the corresponding terms of a converging series \(a_n\), then the series \(\sum b_n\) cannot exceed convergence.

In our problem, we established the inequality: \(0 \leq \frac{\sqrt{a_n}}{n} \leq \frac{a_n}{n}\). This is essential because it sets up the larger series \(\sum \frac{a_n}{n}\) which we intend to show is convergent, thereby ensuring that \(\sum \frac{\sqrt{a_n}}{n}\) converges by the Comparison Test.

This approach can be especially useful when tackling any problem involving positive term series because it only requires comparing it to another known series, providing a clear path to prove convergence without directly calculating complex limits or integrals.

Integral Test

The Integral Test is another potent technique in series convergence analysis, typically applicable for positive term series. It draws a direct link between a series and an integral. If the terms of a series can be represented by a continuous, positive, and decreasing function \(f(x)\) on \(x \geq 1\), then the series \(\sum_{n=1}^\infty f(n)\) converges if and only if the improper integral \[\int_1^\infty f(x)\, dx\] converges.

In our case, consider converting the series \(\sum \frac{a_n}{n}\) into an integral. We need to ensure the corresponding function, \(f(x) = \frac{a_x}{x}\), fits the criteria: positive, continuous, and decreasing. Once we know it's in this form, calculating the integral can directly point to the behavior of the series.

Through calculating \[\int_1^\infty \frac{a_x}{x} \, dx\], we confirm convergence, which helps in the application of the Comparison Test. So, the Integral Test doesn't just stand alone; it forms a bridge enabling the use of other tests, illustrating the interconnectedness of convergence tests.

Series Inequality

Understanding and using inequalities within series is vital, and is not only limited to the Comparison Test setting. In series analysis, series inequality refers to the logical conclusions we can draw from established inequalities, usually to bring an unknown series under the view of already studied or proven behaviors.

In our problem scenario, the inequality \(0 \leq \frac{\sqrt{a_n}}{n} \leq \frac{a_n}{n}\) lets us infer important properties of \(\sum \frac{\sqrt{a_n}}{n}\) by comparing it to \(\sum \frac{a_n}{n}\). Establishing such an inequality is incredibly useful as it directly helps apply the Comparison Test.

The ability to construct valid inequalities often requires a good understanding of the behavior and properties of the functions involved. Recognizing that \(\sqrt{a_n} \leq a_n\) holds due to the nature of the square root and non-negative functions forms the basis of our series inequality here. This groundwork allows us to approach the problem logically and reach conclusions about the series' convergence more effectively.