/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Determine whether the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the following series converge. $$\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k !}{k^{k}}$$

Short Answer

Expert verified
Answer: Yes, the series converges.

Step by step solution

01

The general term of the given series is \(a_k=(-1)^{k+1} \frac{k !}{k^{k}}.\) We will work with this term while checking the convergence conditions. #Step 2: Check if the absolute value of the terms decreases#

To check if the absolute value of the terms decreases, we need to find the ratio \(\frac{|a_{k+1}|}{|a_k|}\) and see if it is less than or equal to 1 for all k. $$\frac{|a_{k+1}|}{|a_k|}=\frac{\left|\frac{(k+1)!}{(k+1)^{k+1}}\right|}{\left|\frac{k!}{k^k}\right|}=\frac{(k+1)!}{k!}\cdot\frac{k^k}{(k+1)^{k+1}}=\frac{k+1}{k}\cdot\left(\frac{k}{k+1}\right)^{k+1}$$ $$\frac{|a_{k+1}|}{|a_k|}=\frac{k+1}{k}\cdot\left(\frac{k}{k+1}\right)^{k+1}=\left(\frac{k}{k+1}\right)^k$$ Now we need to check if \(\lim_{k \to \infty} \left(\frac{k}{k+1}\right)^k \leq 1\). Indeed, the limit exists and equals 1, so the absolute value of the terms decreases for all k. #Step 3: Check if the limit of the terms is 0#
02

Now, let's find the limit of the series terms as k approaches infinity: $$\lim_{k \to \infty} a_k = \lim_{k \to \infty} \left((-1)^{k+1} \frac{k !}{k^{k}}\right)$$ Since the factorial grows faster than the exponential, we have: $$\lim_{k \to \infty} \frac{k !}{k^{k}} =0$$ Thus, the limit of the terms as k approaches infinity is zero. #Step 4: Conclude convergence by the Alternating Series Test#

Since both conditions of the Alternating Series Test are satisfied (the absolute value of the terms decreases and the limit of the terms as k approaches infinity is zero), we can conclude that the given series converges: $$\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k !}{k^{k}}\text{ converges}.$$

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Key Concepts

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

series convergence
A key aspect of series in calculus involves determining whether a series converges or diverges. A series is simply the sum of a sequence of numbers, and we seek to find out if the sum approaches a specific value as we continue adding more terms. This process is known as series convergence.
To establish if a series converges, we often look for tests or methods, such as the Alternating Series Test or the Limit Comparison Test. These tests use specific conditions to systematically check the behavior of the series' terms as they progress toward infinity.
For the series in our exercise, the Alternating Series Test was used. This test focuses on series where the terms alternate between positive and negative. It requires us to establish that:
  • The absolute value of the terms decreases continuously.
  • The limit of the series terms approaches zero as the number of terms becomes infinitely large.
If both these conditions are met, as determined in our step-by-step solution, we can ascertain the series converges.
absolute value of terms
When analyzing a series, particularly for convergence, one essential factor is the behavior of the absolute value of its terms. The absolute value of terms refers to considering each term without regard to its sign, focusing instead on its magnitude. This becomes crucial with alternating series, like the one given in our exercise, where terms switch signs.
Ensuring that the absolute value of terms decreases typically forms a part of convergence tests, such as the Alternating Series Test. The step-by-step solution showed this by computing the limit of the ratio of successive absolute terms:
  • This is expressed as \(\frac{|a_{k+1}|}{|a_k|}\) which needs verification that it is less than or equal to 1 as \(k\) grows.
  • Through mathematical simplification, the example demonstrated that it remained \(\left(\frac{k}{k+1}\right)^k\), which approaches 1.
This conclusion confirms that the absolute values of the terms do decrease, satisfying the first condition for convergence.
limit comparison test
The Limit Comparison Test provides a convenient method to decide if a given series converges by comparing it to another series whose convergence properties are already known. This test contrasts the behavior of two series in terms of their limit.
Although the test was not directly applied in this exercise's solution, understanding its principles enriches the toolkit for series convergence problems. Here's how the test typically works:
  • We take the limit of the ratio of the terms from two series, say \(c_k\) and \(b_k\), in the form \(\lim_{k \to \infty} \frac{c_k}{b_k}\).
  • If this limit is finite and positive, both series will either converge or diverge together.
In practice, you choose a companion series \(b_k\) that resembles \(c_k\)'s behavior but whose convergence is easier to judge. While not used in our specific exercise, the Limit Comparison Test is a versatile tool in series analysis.

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Most popular questions from this chapter

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=2 a_{n}\left(1-a_{n}\right) ; a_{0}=0.3$$

The Greeks solved several calculus problems almost 2000 years before the discovery of calculus. One example is Archimedes' calculation of the area of the region \(R\) bounded by a segment of a parabola, which he did using the "method of exhaustion." As shown in the figure, the idea was to fill \(R\) with an infinite sequence of triangles. Archimedes began with an isosceles triangle inscribed in the parabola, with area \(A_{1}\), and proceeded in stages, with the number of new triangles doubling at each stage. He was able to show (the key to the solution) that at each stage, the area of a new triangle is \(\frac{1}{8}\) of the area of a triangle at the previous stage; for example, \(A_{2}=\frac{1}{8} A_{1},\) and so forth. Show, as Archimedes did, that the area of \(R\) is \(\frac{4}{3}\) times the area of \(A_{1}\).

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{4^{n}+5 n !}{n !+2^{n}}$$

Consider the alternating series $$ \sum_{k=1}^{\infty}(-1)^{k+1} a_{k}, \text { where } a_{k}=\left\\{\begin{array}{cl} \frac{4}{k+1}, & \text { if } k \text { is odd } \\ \frac{2}{k}, & \text { if } k \text { is even } \end{array}\right. $$ a. Write out the first ten terms of the series, group them in pairs, and show that the even partial sums of the series form the (divergent) harmonic series. b. Show that \(\lim _{k \rightarrow \infty} a_{k}=0\) c. Explain why the series diverges even though the terms of the series approach zero.

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n^{1.001} \text { and } b_{n}=\ln n^{10}, n \geq 1$$

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