Convergence Analysis Of A Series With Integral Terms

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Determining the convergence of an infinite series is a fundamental problem in mathematical analysis. This article delves into the convergence analysis of a specific series where the general term is defined by a definite integral. We will explore the techniques and reasoning involved in assessing such series, providing a comprehensive understanding of the process.

Problem Statement

Consider the series $\sum\limits_{n=1}^{\infty}a_n$, where the general term $a_n$ is given by the definite integral:

$a_n=\int\limits_0^{\frac{1}{n}} \dfrac{\sqrt{x}}{1+x^4}dx$, for $n\in\mathbb{N}$.

The objective is to determine whether this series converges or diverges. This problem intricately combines concepts from real analysis, integration, sequences and series, limits, and definite integrals, making it a valuable exercise in mathematical reasoning.

Initial Approaches and Challenges

When faced with such a problem, a natural first step is to try and evaluate the limit of the general term $a_n$ as $n$ approaches infinity. This can provide a preliminary indication of the series' behavior. If $\lim_{n \to \infty} a_n \neq 0$, the series diverges by the Divergence Test. However, if the limit is 0, further investigation is required to determine convergence.

Calculating the limit directly involves evaluating the integral, which might not have a simple closed-form solution. Therefore, we need to explore alternative methods to estimate or bound the integral. This often involves using inequalities and properties of integrals to compare the given integral with simpler functions.

Estimating the Integral Term

Bounding the Integrand

The core idea here is to find suitable upper and lower bounds for the integrand $\dfrac{\sqrt{x}}{1+x^4}$ on the interval of integration $\left[0, \frac{1}{n}\right]$. Since $x^4$ is non-negative, we have $1 + x^4 \geq 1$. Therefore,

$\dfrac{\sqrt{x}}{1+x^4} \leq \sqrt{x}$ for all $x \in [0, \infty)$.

This inequality provides an upper bound that simplifies the integral, making it easier to evaluate. We can also establish a lower bound, though it might not be necessary for determining convergence in this case.

Evaluating the Bounding Integral

Using the upper bound, we can now compare the original integral with a simpler integral:

$0 \leq a_n = \int\limits_0^{\frac{1}{n}} \dfrac{\sqrt{x}}{1+x^4}dx \leq \int\limits_0^{\frac{1}{n}} \sqrt{x} dx$.

The integral on the right-hand side is straightforward to compute:

$\int\limits_0^{\frac{1}{n}} \sqrt{x} dx = \int\limits_0^{\frac{1}{n}} x^{\frac{1}{2}} dx = \left[\dfrac{2}{3}x^{\frac{3}{2}}\right]_0^{\frac{1}{n}} = \dfrac{2}{3}\left(\dfrac{1}{n}\right)^{\frac{3}{2}} = \dfrac{2}{3n^{\frac{3}{2}}}$.

This result gives us an upper bound for $a_n$:

$0 \leq a_n \leq \dfrac{2}{3n^{\frac{3}{2}}}$.

Convergence Analysis using Comparison Test

Applying the Comparison Test

Now that we have an upper bound for $a_n$, we can use the Comparison Test to determine the convergence of the series. The Comparison Test states that if $0 \leq a_n \leq b_n$ for all $n$ and the series $\sum\limits_{n=1}^{\infty} b_n$ converges, then the series $\sum\limits_{n=1}^{\infty} a_n$ also converges.

In our case, we have $0 \leq a_n \leq \dfrac{2}{3n^{\frac{3}{2}}}$. So, we need to analyze the convergence of the series $\sum\limits_{n=1}^{\infty} \dfrac{2}{3n^{\frac{3}{2}}}$.

Recognizing the p-series

The series $\sum\limits_{n=1}^{\infty} \dfrac{2}{3n^{\frac{3}{2}}}$ is a constant multiple of the p-series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{\frac{3}{2}}}$. A p-series has the form $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^p}$, and it converges if $p > 1$ and diverges if $p \leq 1$. In our case, $p = \frac{3}{2}$, which is greater than 1. Therefore, the p-series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{\frac{3}{2}}}$ converges.

Concluding Convergence

Since the p-series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{\frac{3}{2}}}$ converges, the series $\sum\limits_{n=1}^{\infty} \dfrac{2}{3n^{\frac{3}{2}}}$ also converges (as it is a constant multiple of a convergent series). By the Comparison Test, the series $\sum\limits_{n=1}^{\infty} a_n$ converges because $0 \leq a_n \leq \dfrac{2}{3n^{\frac{3}{2}}}$ and $\sum\limits_{n=1}^{\infty} \dfrac{2}{3n^{\frac{3}{2}}}$ converges.

Alternative Approaches and Refinements

Limit Comparison Test

Another approach to this problem is using the Limit Comparison Test. This test compares the limit of the ratio of two series' terms to determine their convergence behavior. If we have two series $\sum a_n$ and $\sum b_n$ with positive terms, and if

$\lim_{n \to \infty} \dfrac{a_n}{b_n} = c$,

where $0 < c < \infty$, then both series either converge or diverge. In our case, we can compare $a_n$ with $b_n = \dfrac{1}{n^{\frac{3}{2}}}$:

$\lim_{n \to \infty} \dfrac{a_n}{\frac{1}{n^{\frac{3}{2}}}}=\lim_{n \to \infty} \dfrac{\int\limits_0^{\frac{1}{n}} \dfrac{\sqrt{x}}{1+x^{4}dx}{\frac{1}{n}{\frac{3}{2}}}} $.

This limit can be evaluated using L'Hôpital's Rule after expressing it in a suitable form. This method provides an alternative route to confirming convergence.

Taylor Series Expansion

While not the most straightforward approach, one could also consider using Taylor series expansions to approximate the integrand. However, this would likely lead to more complex calculations without providing significant advantages in this specific case.

Conclusion

In conclusion, the series $\sum\limits_{n=1}^{\infty}a_n$ with $a_n=\int\limits_0^{\frac{1}{n}} \dfrac{\sqrt{x}}{1+x^4}dx$ converges. The convergence was established by using the Comparison Test, where the integral term was bounded above by a simpler function, allowing comparison with a convergent p-series. This problem showcases the interplay between integration techniques, series convergence tests, and the power of bounding arguments in mathematical analysis. Through a careful examination of the integral and the application of appropriate convergence tests, we successfully determined the behavior of the given series. This exercise reinforces the importance of strategic problem-solving in mathematical analysis, where choosing the right approach can significantly simplify the solution process. The Comparison Test proved to be particularly effective in this scenario, leveraging the properties of p-series to establish convergence. Understanding and applying these techniques are crucial for tackling more complex problems in real analysis and related fields. Moreover, exploring alternative approaches such as the Limit Comparison Test provides a deeper understanding of the problem and its solution landscape. Ultimately, this analysis underscores the rich connections between different areas of mathematics and the power of combining these concepts to solve intricate problems.