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Chapter 9: Problem 37

Use any method to determine whether the series converges. \(\sum_{k=1}^{\infty} \frac{1}{1+\sqrt{k}}\)

Short Answer

Expert verified
The series diverges by the comparison test.

Step by step solution

01

Identify the Type of Series

The given series is \(\sum_{k=1}^{\infty} \frac{1}{1+\sqrt{k}}\). This series involves terms of the form \(\frac{1}{1+\sqrt{k}}\), and as \(k\) increases, this expression can be simplified or compared to a known type, such as a p-series or using the limit comparison test.
02

Use Comparison Test

To use the comparison test, we need to identify a simpler series that is known to converge or diverge and compare it to our series. Consider the series \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\), which is a p-series with \(p=1/2\) and is known to diverge.
03

Compare Terms of the Series

Notice that \( \frac{1}{1+\sqrt{k}} > \frac{1}{2\sqrt{k}} \) for \( k \geq 1 \). Since \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) diverges, by the comparison test \(\sum_{k=1}^{\infty} \frac{1}{2\sqrt{k}}\) diverges as well. Thus, \(\sum_{k=1}^{\infty} \frac{1}{1+\sqrt{k}}\) also diverges.

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Key Concepts

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

Understanding the Comparison Test
The comparison test is a handy tool in your mathematical toolkit when dealing with the convergence of series. It allows you to determine the behavior of a complex series by comparing it to another series whose convergence properties are already known. Here鈥檚 how it works:
  • If you have a series \sum_{k=1}^{\infty} a_k\ that you want to analyze, find another series \sum_{k=1}^{\infty} b_k\ that resembles \sum a_k\ and whose behavior is known.
  • Ensure all terms are positive for both series in your range of interest (typically \(k \geq n\)).
  • If \(a_k \leq b_k\) and \sum b_k\ is known to converge, then \sum a_k\ also converges.
  • If \(a_k \geq b_k\) and \sum b_k\ is known to diverge, then \sum a_k\ also diverges.
The comparison test is most powerful because it connects unknown series to well-understood ones, making it easier to determine convergence or divergence.
Delving into p-Series
A \(\textit{p-series}\) is a series of the form \sum_{k=1}^{\infty} \frac{1}{k^p}\, where \p\ is a positive constant.
  • The behavior of a p-series depends critically on the value of \(p\).
  • When \(p \leq 1\), the series diverges. This means it does not settle into a fixed value but keeps growing.
  • When \(p > 1\), the series converges, allowing it to sum up to a finite number.
In the given exercise, the comparison was made with the series \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\.
This is a p-series with \(p = \frac{1}{2}\), which clearly falls into the diverging category. Because of the divergent behavior of \sum \frac{1}{\sqrt{k}}\, it was a perfect candidate for comparison to predict the behavior of the original problem's series.
Exploring Divergence
Divergence is when the sum of a series doesn't converge to a finite limit. It might grow indefinitely or oscillate without approaching a stable value.
  • In mathematical terms, if \sum_{k=1}^{\infty} a_k\ diverges, the partial sums \S_N = a_1 + a_2 + \cdots + a_N\ do not settle to a specific number as \(N\) approaches infinity.
  • Divergence often requires attention because it signifies that the sum of an infinite number of terms doesn't produce a finite result. This leads to realizing whether further attention or correction is needed during application.
In our example, using the comparison test, it was shown that because the comparison series diverged (the p-series with \(p = \frac{1}{2}\)), the series \sum_{k=1}^{\infty} \frac{1}{1+\sqrt{k}}\ also diverges. Understanding this helps solidify your grasp of when and why certain series are divergent.

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Most popular questions from this chapter

Find the first four nonzero terms of the Maclaurin series for the function by dividing appropriate Maclaurin series. (a) \(\sec x \quad\left(=\frac{1}{\cos x}\right)\) (b) \(\frac{\sin x}{e^{x}}\)

Consider the series $$ \sum_{k=1}^{\infty} \frac{(-3)^{k}}{k} x^{k} $$ Determine the intervals of convergence for this series and for the series obtained by integrating this series term by term.

Prove: (a) If \(f\) is an even function, then all odd powers of \(x\) in its Maclaurin series have coefficient \(0 .\) (b) If \(f\) is an odd function, then all even powers of \(x\) in its Maclaurin series have coefficient \(0 .\)

Show that if \(p\) is a positive integer, then the power series $$ \sum_{k=0}^{\infty} \frac{(p k) !}{(k !)^{p}} x^{k} $$ has a radius of convergence of \(1 / p^{p}\).

Find the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in a known Maclaurin series and performing any algebraic operations that are required. State the radius of convergence of the series. (a) \(\frac{x}{x-1}\) (b) \(3 \cosh \left(x^{2}\right)\) (c) \(\frac{x}{(1+2 x)^{3}}\)
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