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Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \tan \left(n^{1 / n}\right)$$

Short Answer

Expert verified
The series diverges by the Divergence Test since \(\tan(n^{1/n})\) approaches \(\tan(1) \neq 0\).

Step by step solution

01

Analyze the general term

We begin by examining the general term of the series: \(a_n = \tan(n^{1/n})\). As \(n\) approaches infinity, \(n^{1/n}\) approaches \(1\), which means that \(\tan(n^{1/n})\) approaches \(\tan(1)\). Since \(\tan(1)\) is a constant, the term \(a_n\) does not approach zero.
02

Apply the Divergence Test

According to the Divergence Test, if the limit of the sequence \(a_n\) as \(n\) approaches infinity is not zero, then the series \(\sum_{n=1}^{\infty} a_n\) diverges. From our analysis in Step 1, since \(a_n\) approaches \(\tan(1)\), which is not zero, the series diverges.

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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 simple yet powerful tool for determining if an infinite series diverges. It examines the behavior of the general term as the term number grows larger, specifically assessing whether the limit is non-zero. If the limit of the general term \(a_n\) is not zero, it implies that the sum of the infinite series \(\sum_{n=1}^{\infty} a_n\) will not settle to a finite value, leading to divergence.
Understanding the Divergence Test boils down to evaluating \(\lim_{{n \to \infty}} a_n\).
  • If \(\lim_{{n \to \infty}} a_n eq 0\), the series definitely diverges.
  • If \(\lim_{{n \to \infty}} a_n = 0\), the test is inconclusive. The series might converge or diverge still, but other tests are needed.

In the example series \(\sum_{n=1}^{\infty} \tan(n^{1/n})\), the general term \(a_n = \tan(n^{1/n})\) approaches \(\tan(1)\) as \(n\) approaches infinity, and because it does not approach zero, we conclude divergence by the Divergence Test.
Series Analysis
Series Analysis involves examining different aspects and behaviors of a series to understand whether it converges or diverges. The purpose is to find a clear picture of the series behavior, often using a variety of mathematical tools or tests.
There are several strategies commonly employed in this analysis:
  • Divergence Test: Check if the limit of the terms is zero. If not, the series diverges.
  • Comparison Tests: Compare with simpler, known series to draw conclusions.
  • Ratio and Root Tests: Particularly useful for series involving powers and factorials.

Returning to our example, the Divergence Test is enough to determine that the series \(\sum_{n=1}^{\infty} \tan(n^{1/n})\) diverges. However, if the test had been inconclusive, these other methods might have been helpful.
General Term Analysis
Understanding the general term of a series is crucial, as it offers a lens into the series' behavior over large numbers. The general term, denoted as \(a_n\), essentially represents each individual piece added into the series. As we know, the behavior of \(\lim_{{n \to \infty}} a_n\) dictates whether the series might converge or diverge.
Let's consider our example series with \(a_n = \tan(n^{1/n})\). Analyzing \(a_n\), we note:
  • As \(n \to \infty\), \(n^{1/n} \to 1\).
  • Consequently, \(\tan(n^{1/n}) \to \tan(1)\), which is a non-zero constant.

This result is pivotal, as it directly influences the outcome of the divergence test. Since \(\tan(n^{1/n})\) approaches a non-zero limit, the infinite sum cannot converge to a finite number.

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Most popular questions from this chapter

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=2}^{\infty} \ln \left(\frac{n+2}{n+1}\right)$$

Which of the series, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)}$$

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty}\left(1-\frac{2}{n}\right)^{n^{2}}$$

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2 .\) Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

a. Use the binomial series and the fact that $$\frac{d}{d x} \sin ^{-1} x=\left(1-x^{2}\right)^{-1 / 2}$$ to generate the first four nonzero terms of the Taylor series for \(\sin ^{-1} x .\) What is the radius of convergence? b. Series for \(\cos ^{-1} x \quad\) Use your result in part (a) to find the first five nonzero terms of the Taylor series for \(\cos ^{-1} x\)

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