Question

Does there exist a number \(p\) such that \(\sum_{n=1}^{\infty} \frac{2^{n}}{n^{p}}\) converges?

Short answer

No, such a number \(p\) does not exist; the series diverges for all \(p\).

Step-by-step solution

Understand the Series

We are given the series \(\sum_{n=1}^{\infty} \frac{2^n}{n^p}\). Our goal is to determine if there is a value for the parameter \(p\) such that this series converges.

Compare with Known Series Types

This series is similar to a geometric series due to the \(2^n\) term but modified by the \(\frac{1}{n^p}\) term. We will compare it against a geometric series to help analyze convergence.

Recall the p-Series Test

The series \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) converges if \(p > 1\). However, here \(2^n\) will dominate as \(n\) increases, so this condition on \(p\) might not be enough.

Apply the Root Test

The Root Test can determine convergence by examining \( \lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n}{n^p}\right|} \). Calculating this gives \( \lim_{n \to \infty} \frac{2}{n^{p/n}} = 2 \) as \( n^{p/n} \rightarrow 1 \) for any constant \(p\).

Analyze the Root Test Result

The Root Test tells us that if the limit is greater than 1, the series diverges. Here, the limit is 2, which indicates divergence for all \(p\).

Conclusion on Convergence

Based on the Root Test, there does not exist any positive \(p\) that makes \(\sum_{n=1}^{\infty} \frac{2^n}{n^p}\) converge, because the limit calculated was 2, which indicates divergence.

Key concepts

p-series

A p-series is a type of mathematical series represented by the expression \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). This series converges when the value of \( p \) is greater than 1. Convergence here means that as you keep adding more terms from this series, the total sum approaches a finite number.With p-series, the value of \( p \) is crucial. If \( p \) is less than or equal to 1, the terms do not get small enough as \( n \) increases, and the series diverges, meaning it does not have a finite sum. In the context of the problem involving the series \( \sum_{n=1}^{\infty} \frac{2^n}{n^p} \), we tried to apply the conditions for convergence of a p-series. However, the term \( 2^n \) complicates this, as it grows exponentially. Thus, the series behaves differently than a standard p-series, especially for larger \( n \).

geometric series

A geometric series is another important type of series that is defined by the expression \( \sum_{n=0}^{\infty} ar^n \). In this formula, \( a \) represents the first term, and \( r \) is called the common ratio. When \( |r| < 1 \), the geometric series converges. If \( |r| \geq 1 \), it diverges. In the given series problem, the presence of \( 2^n \) makes the series appear similar to a geometric series with a common ratio \( r = 2 \), which is greater than 1. This suggests that without any modifying factor, like the \( \frac{1}{n^p} \), the series would diverge. This is because the terms get larger as each subsequent term is multiplied by 2. However, the \( \frac{1}{n^p} \) factor alters this scenario slightly but not sufficiently enough to ensure convergence when \( r = 2 \). So the series behaves much like a geometric series that diverges.

root test

The root test is a method used to determine the convergence of a series. It involves finding the limit \[L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\]where \( a_n \) are the terms of the series. The root test states:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

For the given series \( \sum_{n=1}^{\infty} \frac{2^n}{n^p} \), we applied the root test: \[L = \lim_{n \to \infty} \sqrt[n]{\left|\frac{2^n}{n^p}\right|} = 2\]Since the limit equals 2, which is greater than 1, the series diverges. The root test efficiently shows that despite the factor \( \frac{1}{n^p} \), the dominating exponential growth of \( 2^n \) ensures the series does not converge for any positive value of \( p \).

series divergence

Series divergence means that the sum of an infinite series does not settle to a finite limit. Instead, it grows in magnitude, or oscillates indefinitely without approaching any fixed value. This can occur if the terms of the series do not decrease fast enough, or if there is persistent growth due to factors like geometric scaling.In the given problem with the series \( \sum_{n=1}^{\infty} \frac{2^n}{n^p} \), divergence was concluded when the root test yielded a result of 2, exceeding the threshold of 1 for convergence. The term \( 2^n \) grows exponentially. No matter how large \( p \) is chosen, the exponential growth outpaces the offsetting influence of \( \frac{1}{n^p} \). Thus, the series continues to grow indefinitely, illustrating a case of divergence. Understanding divergence helps in determining the feasibility of a series achieving a sum and acknowledging that some series will naturally expand beyond bounds due to their structure.