Problem 8

Question

Determine whether the series converges or diverges. It is possible to solve Problems 4 through 19 without the Limit Comparison, Ratio, and Root Tests. $\sum_{k=1}^{\infty} \frac{k}{e^{k}}$

Step-by-Step Solution

Verified

Answer

The series $\sum_{k=1}^{\infty} \frac{k}{e^{k}}$ is convergent.

1Step 1: Identify the Series Type

The given series is $\sum_{k=1}^{\infty} \frac{k}{e^{k}}$ and it's a form of p-series combined with a geometric series wherein $k$ represents the 'n' term in the series and $e^{k}$ implies an exponential function.

2Step 2: Apply the Limit Test

The Limit test can be employed here because it's easier to apply and also it's one of the recommended tests to use in the problem statement. The limit of the $n$th term as $n$ goes to infinity is found using the limit test. This is formally given by: $\lim_{k\to\infty} \frac{k}{e^{k}}$. Applying L'Hopital's rule since it is an indeterminate form, we get: $\lim_{k\to\infty} \frac{1}{e^{k}} = 0$.

3Step 3: Conclude Convergence or Divergence

Since the limit of the $n$th term as $n$ goes to infinity is 0, according to the Limit Test, the series $\sum_{k=1}^{\infty} \frac{k}{e^{k}}$ is convergent.

Key Concepts

Understanding p-seriesExploring Geometric SeriesUtilizing the Limit TestApplying L'Hopital's Rule

Understanding p-series

A p-series is a specific type of infinite series given by the formula $\sum_{k=1}^{\infty} \frac{1}{k^p}$, where $p$ is a constant. The convergence or divergence of a p-series relies heavily on the value of $p$.

If $p > 1$, the series converges.
If $p \leq 1$, the series diverges.

In the exercise, the given series $\sum_{k=1}^{\infty} \frac{k}{e^{k}}$ is not exactly a p-series. However, understanding the idea behind p-series can offer insight into approaching series problems. By breaking down series into recognizable components like p-series, we can strategically determine convergence behavior.Understanding this helps when evaluating complex series by analyzing simpler underlying structures.

Exploring Geometric Series

A geometric series is another fundamental type of series characterized by its constant ratio between successive terms. It is typically expressed in the form $\sum_{k=0}^{\infty} ar^k$, where $a$ is the first term and $r$ is the common ratio.

If $|r| < 1$, the series converges to $\frac{a}{1-r}$.
If $|r| \geq 1$, the series diverges.

In the series $\sum_{k=1}^{\infty} \frac{k}{e^{k}}$, the term $e^k$ introduces an exponential decay, much like the common ratio $r$ in a geometric series. This insight helps in recognizing that exponential terms often guide convergence by ensuring that terms of the series shrink rapidly. Knowing the behavior of geometric series is crucial as it provides a foundation for understanding more complex series that may contain exponential components.

Utilizing the Limit Test

The Limit Test, also known as the nth-term test for divergence, is a straightforward criterion used to test the convergence of series. According to the Limit Test, if $\lim_{k \to \infty} a_k eq 0$, the series $\sum a_k$ diverges. However, if $\lim_{k \to \infty} a_k = 0$, the test is inconclusive and we need further analysis.In the exercise, the series was evaluated using this test. The given series $\sum_{k=1}^{\infty} \frac{k}{e^{k}}$ was found to have a limit:\[\lim_{k \to \infty} \frac{k}{e^k} = 0\]Thus, the Limit Test indicates potential convergence, but further investigation with other methods, like applying L'Hopital's Rule, confirmed conclusions about the series' convergence.

Applying L'Hopital's Rule

L'Hopital's Rule is a valuable technique used to evaluate limits that present indeterminate forms, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. To apply L'Hopital's Rule:

Ensure the limit is in an indeterminate form.
Differentiate the numerator and the denominator separately.
Re-evaluate the limit using these derivatives.

In the given exercise, while determining the limit $\lim_{k \to \infty} \frac{k}{e^k}$, it was initially in the form of $\frac{\infty}{\infty}$. Here, L'Hopital's Rule was applied:\[\lim_{k \to \infty} \frac{k}{e^k} = \lim_{k \to \infty} \frac{1}{e^k} = 0\]This validated that the exponential function in the denominator grows significantly faster than the linear function in the numerator, leading to a conclusion that helps us understand the convergence of the entire series. Applying such a method is crucial when tackling complex limits in series convergence analysis.

Problem 7

Problem 8

Other exercises in this chapter

Problem 7

Find the Taylor series for $f(x)$ centered at the indicated value of $b$. $$ f(x)=\frac{1}{\sqrt{x}}, \quad b=1 $$

View solution

Problem 7

Do the following. (a) Compute the fourth degree Taylor polynomial for $f(x)$ at $x=0 .$ (b) On the same set of axes, graph \(f(x), P_{1}(x), P_{2}(x), P_{3}

View solution

Problem 8

For each series, determine whether the series converges absolutely, converges conditionally, or diverges. $$ \sum_{k=1}^{\infty}(-1)^{k} \frac{2^{k}}{k} $$

View solution

Problem 8

The second degree Taylor polynomial generated by $\ln (1+x)$ about $x=0$ is $x-\frac{x^{2}}{2}$. Use Taylor s Theorem to nd a good upper bound on the erro

View solution