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Determine whether the series converges or diverges. \(\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^{2}+1}\)

Short Answer

Expert verified
The series converges by the Comparison Test.

Step by step solution

01

Understanding the Series

The series given is \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^{2}+1} \). We need to determine whether this series converges or diverges. A series converges if the sum of its infinite terms approaches a finite number and diverges otherwise.
02

Choose a Convergence Test

To analyze the convergence of the series, we choose the Comparison Test. This test is useful when the form of the series suggests a simpler series for which we already know the convergence behavior.
03

Compare with a Known Series

We observe that \( \frac{\sqrt{n}}{n^{2}+1} < \frac{\sqrt{n}}{n^{2}} = \frac{1}{n^{3/2}} \) for all \( n \geq 1 \). The series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) is known to converge because it's a p-series with \( p = \frac{3}{2} > 1 \).
04

Apply the Comparison Test

Since \( \frac{\sqrt{n}}{n^{2}+1} < \frac{1}{n^{3/2}} \) and \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) converges, the Comparison Test indicates that \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^{2}+1} \) also converges. Therefore, our original series converges as well.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Convergence
Understanding whether a series converges or diverges is crucial in calculus. A series is a sum of terms in a sequence. The infinite series \( \sum_{n=1}^{\infty} a_n \) converges if its terms \( a_n \) add up to a finite limit as \( n \) approaches infinity. If the sum keeps increasing indefinitely, the series diverges, which means it does not have a finite sum.

To determine the convergence of a series, we employ various tests such as the Comparison Test, Ratio Test, Root Test, and others. These tests help in understanding the behavior of series, especially when dealing with non-obvious sequences.

In the case of power series, where terms involve powers of \( n \), these tests become particularly useful.
Comparison Test
The Comparison Test is a popular method used to determine convergence or divergence of a series by comparing it to another series with known behavior.

The basic idea is simple:
  • Identify a series with known convergence or divergence.
  • Compare each term of the given series with a corresponding term of the known series.
  • If the terms of the given series are smaller than the terms of a convergent known series, the given series also converges.
  • If the terms are larger and the known series diverges, then the given series also diverges.
A successful comparison requires insight into the nature of the given series and the identification of an appropriate comparator. For example, in our original step solution, the series \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2+1} \) is compared to the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \), which is known to converge.
P-series
P-series provide a standard collection of series whose convergence or divergence is well studied and established. A p-series takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant.

The convergence depends on the value of \( p \):
  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
The simplicity of p-series makes them excellent comparators for the Comparison Test.

In the original exercise, understanding that \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) (with \( p = \frac{3}{2} \)) converges helps us infer the behavior of the given series. P-series allow us to apply established rules to more complex series by comparison.

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