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Test for convergence or divergence and identify the test used.\(\sum_{n=1}^{\infty} \frac{3}{n^{2}+5}\)

Short Answer

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The series \(\sum_{n=1}^{\infty} \frac{3}{n^{2}+5}\) converges.

Step by step solution

01

Identify Test To Be Used

The Comparison Test or Limit Comparison Test can be used to determine if the series converges or diverges. The comparator will be \(\frac{1}{n^{2}}\) since it has some similarities with the given series.
02

Application of the Comparison Test

Set up the series alongside the comparator and form a ratio: \( \frac{\frac{3}{n^{2}+5}}{\frac{1}{n^{2}}}\). Simplify the ratio to get \( \frac{3}{1 + \frac{5}{n^{2}}}\). Now, evaluate the limit as n approaches infinity. That is: \( \lim_{n \to \infty} \frac{3}{1 + \frac{5}{n^{2}}}\) which equals 3.
03

Evaluate and Make Conclusion

Since the limit is 3 (a positive finite number), and the comparator series \(\frac{1}{n^{2}}\) is a p-series with \( p = 2 > 1 \) (which is convergent), we can say that the original series \(\sum_{n=1}^{\infty} \frac{3}{n^{2}+5}\) also converges by the Limit Comparison Test.

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Key Concepts

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

Comparison Test
The Comparison Test is a powerful tool in calculus for determining the convergence or divergence of an infinite series. When you have an infinite series, \(\sum_{n=1}^{\infty} a_n\), you compare it to another series with known behavior, \(\sum_{n=1}^{\infty} b_n\), to make a conclusion about the convergence or divergence of \(\sum_{n=1}^{\infty} a_n\).

To use the Comparison Test effectively:
  • Select a comparator series \(\sum_{n=1}^{\infty} b_n\) that resembles the original series you are analyzing.
  • If every term \(a_n \leq b_n\) and \(\sum_{n=1}^{\infty} b_n\) is known to converge, then \(\sum_{n=1}^{\infty} a_n\) will also converge.
  • Conversely, if \(a_n \geq b_n\) and \(\sum_{n=1}^{\infty} b_n\) is known to diverge, then \(\sum_{n=1}^{\infty} a_n\) will also diverge.
Always ensure the series you are comparing to has simpler terms and a clear behavior (convergence or divergence).
Limit Comparison Test
The Limit Comparison Test is similar in nature to the Comparison Test, but it requires taking a limit. It's utilized when the terms of the series you are testing do not lend themselves to a direct comparison.

Here's how you proceed:
  • First, identify a comparator series \(\sum_{n=1}^{\infty} b_n\), that has a similar n-term as the original series \(\sum_{n=1}^{\infty} a_n\).
  • Next, calculate the limit of the ratio of the nth terms: \(\lim_{n \to \infty} \frac{a_n}{b_n}\).
  • If this limit results in a positive finite number, both series will have the same behavior - they will both converge or both diverge.
  • The success of this test crucially relies on the proper choice of a comparator series \(b_n\) which should be a simpler series with known convergence properties.
p-series convergence
In the realm of infinite series, a p-series is expressed as \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), where \(p\) is any positive real number. The convergence of a p-series depends on the value of \(p\):

  • If \(p > 1\), the series converges.
  • If \(p \leq 1\), the series diverges.
This is a fundamental result that provides a benchmark for comparing other more complex series. Knowing the behavior of p-series is incredibly helpful when using either the Comparison Test or the Limit Comparison Test, as they can often serve as a comparator series.
Infinite Series
An infinite series is essentially a sum of an infinite sequence of terms, expressed as \(\sum_{n=1}^{\infty} a_n\). Infinite series can either converge, which means they add up to a specific finite value, or diverge, which means they do not sum to a finite value.

There are multiple tests to determine the behavior of an infinite series, such as: Comparison Test, Limit Comparison Test, Integral Test, Ratio Test, and Root Test, among others. Each test applies under different conditions and may require certain characteristics of the series to be utilized. It is vital to have a good understanding of these tests and to select the appropriate one based on the series in question to draw accurate conclusions regarding their convergence or divergence.

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Most popular questions from this chapter

Let \(f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}\), where \(c_{n+3}=c_{n}\) for \(n \geq 0 .\) (a) Find the interval of convergence of the series. (b) Find an explicit formula for \(f(x)\).

(a) Find the common ratio of the geometric series, (b) write the function that gives the sum of the series, and (c) use a graphing utility to graph the function and the partial sums \(S_{3}\) and \(S_{5^{*}}\). What do you notice? \(1-\frac{x}{2}+\frac{x^{2}}{4}-\frac{x^{3}}{8}+\cdots\)

(a) You delete a finite number of terms from a divergent series. Will the new series still diverge? Explain your reasoning. (b) You add a finite number of terms to a convergent series. Will the new series still converge? Explain your reasoning.

The random variable \(n\) represents the number of units of a product sold per day in a store. The probability distribution of \(n\) is given by \(P(n) .\) Find the probability that two units are sold in a given day \([P(2)]\) and show that \(P(1)+P(2)+P(3)+\cdots=1\) $$ P(n)=\frac{1}{3}\left(\frac{2}{3}\right)^{n} $$

Use the formula for the \(n\) th partial sum of a geometric series \(\sum_{i=0}^{n-1} a r^{i}=\frac{a\left(1-r^{n}\right)}{1-r}\) Annuities When an employee receives a paycheck at the end of each month, \(P\) dollars is invested in a retirement account. These deposits are made each month for \(t\) years and the account earns interest at the annual percentage rate \(r\). If the interest is compounded monthly, the amount \(A\) in the account at the end of \(t\) years is $$ \begin{aligned} A &=P+P\left(1+\frac{r}{12}\right)+\cdots+P\left(1+\frac{r}{12}\right)^{12 t-1} \\ &=P\left(\frac{12}{r}\right)\left[\left(1+\frac{r}{12}\right)^{12 t}-1\right] \end{aligned} $$ If the interest is compounded continuously, the amount \(A\) in the account after \(t\) years is $$ \begin{aligned} A &=P+P e^{r / 12}+P e^{2 r / 12}+P e^{(12 t-1) r / 12} \\ &=\frac{P\left(e^{n}-1\right)}{e^{r / 12}-1} \end{aligned} $$ Verify the formulas for the sums given above.

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