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State whether the given \(p\) -series converges. \(\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n^{4}}}\)

Short Answer

Expert verified
The series converges because \( p = \frac{4}{3} > 1 \).

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n^{4}}} \). This can be rewritten as \( \sum_{n=1}^{\infty} n^{-\frac{4}{3}} \). This is a \( p \)-series where \( p = \frac{4}{3} \).
02

Determine p-series Convergence Condition

A \( p \)-series \( \sum_{n=1}^{\infty} n^{-p} \) converges if and only if \( p > 1 \). If \( p \leq 1 \), the series diverges.
03

Evaluate p Against Convergence Condition

In this case, \( p = \frac{4}{3} \). Since \( \frac{4}{3} > 1 \), the series \( \sum_{n=1}^{\infty} n^{-\frac{4}{3}} \) converges.

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Key Concepts

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

Convergence Condition
Understanding the convergence condition is crucial for determining whether a series converges or not. For a \(p\)-series, which has the general form \(\sum_{n=1}^{\infty} n^{-p}\), the convergence is dependent on the value of \(p\). Specifically, a \(p\)-series converges if \(p > 1\). This condition arises because as \(n\) increases, the terms in the series diminish faster if \(p\) is greater than one, leading the series to sum to a finite number.

However, if \(p \leq 1\), the terms do not diminish fast enough, causing the series to diverge. In our example, where \(p = \frac{4}{3}\), we see that \(\frac{4}{3} > 1\), which is why our series converges.

The convergence condition is a straightforward yet powerful concept, enabling determination of convergence by simply comparing the \(p\) value with 1.
Series Type
Series type in mathematics refers to the form or pattern in which the terms of a sequence are arranged. Recognizing the type of a series is essential for determining its properties, such as convergence. One common series type is the \(p\)-series, characterized by terms of the form \(n^{-p}\).

In our example, the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n^{4}}}\) was initially unclear, but by rewriting it as \(\sum_{n=1}^{\infty} n^{-\frac{4}{3}}\), we clearly identified it as a \(p\)-series.

In mathematical analysis, identifying the series type accurately is like using a guidebook; it directs you to the correct methods and conditions needed to analyze convergence or divergence. Recognizing a \(p\)-series allows us to immediately apply the relevant convergence tests.
Educational Mathematics
Educational mathematics encompasses the methods and approaches used to teach mathematical concepts, ensuring students grasp complex topics successfully. For the \(p\)-series, breaking the problem down into steps is an effective teaching method.

Firstly, identify the type of series; this sets the stage for further evaluation. Next, apply convergence conditions specific to the series type, reinforcing understanding through practical application. Finally, conclude by evaluating the derived parameter (\(p\)) to determine convergence.

Using these structured steps not only aids in solving the problem but also solidifies the concept for students. This approach ensures that learners not only memorize formulas but deeply understand how and why they work, cultivating long-term mathematical competence.

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