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Determine whether the given sequence converges or diverges and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$a_{n}=\frac{1}{\sqrt[3]{n}}+\frac{1}{\sqrt[n]{3}}$$

Short Answer

Expert verified
The sequence converges to 1.

Step by step solution

01

Identify the Limits Involved

Begin by considering the behavior of each term in the sequence as \(n\) approaches infinity. The sequence is given \(a_n = \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[n]{3}}\). Evaluate the limit of \(\frac{1}{\sqrt[3]{n}}\) and \(\frac{1}{\sqrt[n]{3}}\) separately as \(n\) goes to infinity.
02

Evaluate the First Term

For the first term \(\frac{1}{\sqrt[3]{n}}\), rewrite it as \(\frac{1}{n^{1/3}}\). As \(n\) approaches infinity, the power of \(n\), which is \(-1/3\), implies this term tends towards 0.
03

Evaluate the Second Term

For the second term \(\frac{1}{\sqrt[n]{3}}\), note that the expression \(\sqrt[n]{3}\) can be written as \(3^{1/n}\). As \(n\) approaches infinity, \(3^{1/n}\) approaches 1 (because \(x^{1/n} \rightarrow 1\) for any positive \(x\) as \(n\) approaches infinity). Therefore, \(\frac{1}{3^{1/n}} \rightarrow \frac{1}{1} = 1\).
04

Combine the Results

Now combine the results of the two evaluated terms. The limit of the sequence \(a_n\) is given by adding the limits of the two individual terms: \(0 + 1 = 1\).
05

Conclude the Convergence

Since both terms have finite limits and their sum is finite, the sequence \(a_n\) converges. Therefore, the sequence converges to the limit 1.

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Key Concepts

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

Limits
Limits are essential in analyzing sequences and their behaviors as they reach infinity. When evaluating limits, we seek the value that the terms of a sequence approach as we consider more terms.
In our given sequence, denoted by \(a_n = \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[n]{3}}\), two separate expressions must be considered.
  • The first expression, \(\frac{1}{\sqrt[3]{n}}\), changes form to \(\frac{1}{n^{1/3}}\). As \(n\) grows larger and larger, the expression tends to zero due to the negative exponent \(-1/3\).
  • The second expression, \(\frac{1}{\sqrt[n]{3}}\), as rewritten as \(3^{1/n}\), approaches one for large \(n\). Thus, its reciprocal moves towards 1.
Examining each limit helps confirm the overall limit of the sequence, helping us determine whether it converges. A sequence converges if its terms tend toward a specific number as \(n\) goes to infinity.
Infinity
Infinity in mathematical terms represents the concept of boundlessness. When we say a sequence behaves 'as \(n\) approaches infinity,' we mean we observe the terms as they progress without end.
In sequences, this concept is fundamental. As \(n\rightarrow\infty\), the behavior of each term in \(a_n\), our sequence, determines if it eventually nears a finite limit.
The first term, \(\frac{1}{\sqrt[3]{n}}\), diminishes with increasing \(n\), driven by the growing denominator making the entire fraction small. The second term, \(\frac{1}{\sqrt[n]{3}}\), converges to 1, illustrating how the concept of infinity aids in understanding behavior over an infinite span.
These observations help us resolve whether the series' sum is finite or not, gauging its convergence based on the behavior seen when \(n\) reaches infinity.
Terms Evaluation
Terms evaluation in sequences involves analyzing each part independently to comprehend the overall behavior. For \(a_n\), this evaluation occurs for the terms \(\frac{1}{\sqrt[3]{n}}\) and \(\frac{1}{\sqrt[n]{3}}\). Let's take a closer look:
The first term transforms to \(\frac{1}{n^{1/3}}\). As previously outlined, raising \(n\) to an increasingly larger power in the denominator ensures this term trends toward zero.
The second term, adjusted to \(3^{1/n}\), becomes more complex, requiring recognition that as the size of \(n\) increases, the power of the base reduces to zero, making \(3^{1/n}\) head towards 1.
The evaluation combines the behavior: \(0 + 1 = 1\).
  • Each term's individual limit contributes to the sequence's fate, either as convergent or divergent.
  • Through steady analysis, it quickly becomes clear whether the sequence converges to a specific number or not.
Applying step-by-step term evaluation greatly clarifies convergence and the sequence's long-term behavior.

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Most popular questions from this chapter

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\frac{1}{1-x} \cosh x $$

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Let $$ f(x)=\left\\{\begin{array}{ll} e^{-1 / x^{2}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0 \end{array}\right. $$ (a) Show that \(f^{\prime}(0)=0\) by using the definition of the derivative. (b) Show that \(f^{\prime \prime}(0)=0\). (c) Assuming the known fact that \(f^{(n)}(0)=0\) for all \(n,\) find the Maclaurin series for \(f(x)\) (d) Does the Maclaurin series represent \(f(x)\) ? (e) When \(a=0,\) the formula in Theorem \(\mathrm{B}\) is called Maclaurin's Formula. What is the remainder in Maclaurin's Formula for \(f(x) ?\) This shows that a Maclaurin series may exist and yet not represent the given function (the remainder does not tend to 0 as \(n \rightarrow \infty)\)

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