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Determine whether the series converges. $$\sum_{k=1}^{\infty} \frac{1}{\sqrt[k]{e}}$$

Short Answer

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The series \( \sum_{k=1}^{\infty} \frac{1}{\sqrt[k]{e}} \) diverges.

Step by step solution

01

Understand the Form of the Series

The series given is \( \sum_{k=1}^{\infty} \frac{1}{\sqrt[k]{e}} \), which can be rewritten as \( \sum_{k=1}^{\infty} e^{-1/k} \) because \( \sqrt[k]{e} \) is equivalent to \( e^{1/k} \).
02

Analyze the General Term

The general term of the series is \( a_k = e^{-1/k} \). Notice that as \( k \) approaches infinity, \( -1/k \) approaches zero which means \( e^{-1/k} \) approaches 1. Hence, \( a_k \) approximates to 1 for large \( k \).
03

Apply the Divergence Test

Since the general term \( a_k = e^{-1/k} \) does not approach 0 as \( k \to \infty \) (it approaches 1), the series cannot converge according to the divergence test. For a series \( \sum a_k \) to converge, \( \lim_{k \to \infty} a_k \) must be 0.

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Key Concepts

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

Divergence Test
When analyzing whether a series converges or not, the Divergence Test, also known as the nth-Term Test for Divergence, plays a crucial role. If the series is represented as \( \sum a_k \), to apply this test, we first look at the limit of its general term \( a_k \). The key here is:
  • If \( \lim_{k \to \infty} a_k eq 0 \), the series diverges.
  • If \( \lim_{k \to \infty} a_k = 0 \), the test is inconclusive (meaning the series might still converge or diverge).
In the original problem, the series \( \sum_{k=1}^{\infty} e^{-1/k} \) was analyzed, where the term \( a_k = e^{-1/k} \) approaches 1 as \( k \) approaches infinity. Because the term does not approach zero, the series diverges. It's important to note that the divergence test is only a preliminary check. Failing this test (when \( \lim_{k \to \infty} a_k = 0 \)) requires further testing, as it does not guarantee convergence.
Limit of a Sequence
Understanding how the limit of a sequence works is very important when dealing with series convergence. A sequence \((a_k)\) has a limit \( L \) if, as \( k \) becomes very large, \( a_k \) gets arbitrarily close to \( L \). In mathematical terms, this is written as:
  • \( \lim_{k \to \infty} a_k = L \)
In the context of series, particularly when considering the divergence test, identifying whether \( a_k \) tends towards zero is vital for the possibility of convergence.
In our exercise, the sequence formed by the general terms, \( e^{-1/k} \), approached the limit 1 and not 0, implying divergence. Calculating the limit of a sequence often involves techniques like L'Hôpital's Rule or considering dominant terms that simplify calculations. Mastering limits is foundational for analyzing series behavior.
Exponential Functions
Exponential functions are a common encounter in series, particularly in formulas involving growth or decay. An exponential function has the form \( f(x) = a^{x} \) or the natural exponential using the base \( e \), \( f(x) = e^{x} \). In our problem, we see the term \( e^{-1/k} \). It represents an exponential decay where the power approaches zero as \( k \) goes to infinity.
  • Exponentials can drastically change the behavior of functions and sequences.
  • Understanding their properties, such as swift growth or decay rates, helps in determining limits and behavior of series.
The natural base \( e \) is approximately 2.718 and is special due to properties in continuously compounding contexts. Recognizing how terms like \( e^{-1/k} \) simplify to familiar forms (e.g., they get closer to 1) as \( k \) becomes infinitely large allows us to utilize them effectively in calculations, gauging how sequences and series behave analytically.

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