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Chapter 9: Problem 11

Determine whether the given series must diverge because its terms do not converge to \(0 .\) $$ \sum_{n=1}^{\infty} \sin \left(\frac{\pi}{2}-\frac{1}{n}\right) $$

Short Answer

Expert verified
The series diverges because its terms do not converge to 0.

Step by step solution

01

Analyze the Term Behavior

Consider the term given by the series: \( a_n = \sin\left(\frac{\pi}{2} - \frac{1}{n}\right) \). Recognize that \( \frac{\pi}{2} \) suggests an approach to a known angle where sine has a special value.
02

Use Sine Angle Approximation

Apply the trigonometric identity for small angles: \( \sin\left(\frac{\pi}{2} - x\right) = \cos(x) \) for small \( x \). Since \( x = \frac{1}{n} \) is small as \( n \to \infty \), we have \( a_n \approx \cos\left(\frac{1}{n}\right) \).
03

Evaluate Limit of Cosine Function

Determine the limit: \( \lim_{n \to \infty} \cos\left(\frac{1}{n}\right) = \cos(0) = 1 \). Since the cosine function approaches 1 as \( n \to \infty \), the terms of the series do not converge to 0.
04

Apply Divergence Test

Since the limit of the terms \( a_n \) is not 0 (it is 1), by the divergence test, the series \( \sum_{n=1}^{\infty} a_n \) must diverge.

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Key Concepts

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

Divergence Test
The Divergence Test is a fundamental tool you can use to quickly determine whether a series diverges. It helps you establish if the infinite sum of a sequence has no finite limit. Here's how it works:
  • If the limit of the sequence's terms, as they approach infinity, is not zero, then the series must diverge. This means \[\lim_{{n \to \infty}} a_n eq 0\] will automatically imply the divergence of \(\sum a_n\).
  • On the other hand, if the terms do approach zero, the Divergence Test does not provide any information about convergence and you must use other methods to investigate further.
In our problem, the term \(a_n\), involving a sine expression, was found to have a non-zero limit. Therefore, using the Divergence Test, we concluded that the series diverges.
Trigonometric Identities
Trigonometric identities are useful tools in simplification and analysis of functions and sequences, especially when they include angles. In this problem, we dealt with a series involving the sine function.
  • A crucial identity used here is that for small angles: \[\sin\left(\frac{\pi}{2} - x\right) = \cos(x)\]
  • This identity is particularly handy when \(x\) is very small, as in the case of \(x = \frac{1}{n}\), helping us approach the behavior of the original sine expression.
By transforming the sine to cosine, you simplify the problem to a function (\(\cos(x)\)) that is easier to analyze, especially for small angles, which allowed us to assess the convergence more directly.
Limit of a Function
Finding the limit of a function as it tends toward a certain point is a cornerstone concept in calculus. When you evaluate a limit, you're essentially trying to determine what value a function approaches as the input approaches a certain number. For infinite series, you want to find the limit of the individual terms.
  • For the function \(a_n = \cos\left(\frac{1}{n}\right)\), you examined what happens as \(n\) grows larger and larger.
  • In this case, as \(n \to \infty\), \(\frac{1}{n}\) approaches 0, leading to \(\cos(0) = 1\).
Knowing this, the terms \(a_n\) do not approach zero, leading to the series' divergence. This limit evaluation showed that the Divergence Test applies, confirming the divergence of the original series.

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Most popular questions from this chapter

Let \(p>1\). Show that for any integer \(j\), the \(j\) th truncation error \(E_{j}\) for \(\sum_{n=1}^{\infty} 1 / n^{p}\) satisfies the inequalities $$ \frac{1}{(p-1)(j+1)^{p-1}} \leq E_{j} \leq \frac{1}{(p-1) j^{p-1}} $$

Approximate the value of the integral with an error less than the given error, first using the Integration Theorem to express the integral as an infinite series and then approximating the infinite series by an appropriate partial sum. $$ \int_{0}^{1} \cos x^{2} d x ; 10^{-7} $$

An indeterminately large number of identical blocks 1 unit long are stacked on top of each other. Show that it is possible for the top block to protrude as far from the bottom block as we wish without the blocks toppling (Figure 9.18). (Hint: The center of gravity of the top block must lie over the second block; the center of gravity of the top two blocks must lie over the third block, and so on. Thus the top block can protrude up to \(\frac{1}{2}\) unit from the end of the second block, the second block can protrude up to \(\frac{1}{4}\) unit from the end of the third block, the third block can protrude up to \(\frac{1}{6}\) unit from the end of the fourth block, and so on. Assuming that the center of gravity of the first \((n-1)\) blocks lies over the end of the \(n\) th block, show that the \(n\) th block can protrude up to \(1 / 2 n\) units from the end of the \((n+1)\) st block.)

Find the Taylor series of \(f\) about \(a\), and write out the first four terms of the series. $$ f(x)=\frac{x}{\sqrt{1-x^{2}}} ; a=0 $$

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n !}{100^{n}} $$
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