/*! 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 31 Determine the convergence or div... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the convergence or divergence of the following series. $$\sum_{k=3}^{\infty} \frac{1}{(k-2)^{4}}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Identify a comparable series

We are looking to compare our given series to another series that we know converges or diverges. Consider the p-series, which has the form: $$\sum_{n=1}^{\infty} \frac{1}{n^p}$$ This converges when p > 1 and diverges when p ≤ 1. In our case, the given series has a similar form to the p-series when p = 4, which satisfies the condition p > 1.
02

Apply the Limit Comparison Test

We will compare the given series with the convergent p-series (where p = 4). We will find the limit as k approaches infinity of the ratio of the terms in the two series: $$\lim_{k\to\infty} \frac{\frac{1}{k^4}}{\frac{1}{(k-2)^4}}$$
03

Calculate the limit

We can simplify the expression by multiplying both the numerator and the denominator by \((k-2)^{4}\): $$\lim_{k\to\infty} \frac{(k-2)^4}{k^4}$$ Now, let's divide the numerator and denominator by \(k^4\): $$\lim_{k\to\infty} \frac{((-2)^4k^0 + 0k^1 - 0k^2 + 0k^3 + k^4)}{k^4}$$ Notice that the highest powers of k in both the numerator and the denominator are equal (both are \(k^4\)). This means that the limit of the ratio of the leading coefficients exists and is finite: $$\lim_{k\to\infty} \frac{1}{1} = 1$$ Since the limit is finite and greater than zero, by the Limit Comparison Test, the given series converges.
04

Conclusion

The original series, \(\sum_{k=3}^{\infty} \frac{1}{(k-2)^{4}}\), converges by the Limit Comparison Test.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a powerful tool in calculus that helps us determine whether a series converges or diverges. It is particularly useful when a series resembles a simpler or well-known series, like a p-series.
This test involves taking the limit of the ratio of the terms of the given series and a known series as the variable approaches infinity. If the limit is a finite number greater than zero, both series will either converge or diverge together.
For example, if you have a series \ \( \sum_{k=1}^{\infty} a_k \ \) and you are comparing it to another series \ \( \sum_{k=1}^{\infty} b_k \ \), you calculate \ \( \lim_{k \to \infty} \frac{a_k}{b_k} \ \). If this limit is a positive, finite number, then both series are either both converging or both diverging.
This is very helpful because if we can find a comparable series that we know converges, we can conclude the same about our original series. Likewise, if the known series diverges, the original series does too.
P-series
A p-series is a specific type of infinite series with the general form \ \( \sum_{n=1}^{\infty} \frac{1}{n^p} \ \). The determination of a p-series's convergence is entirely dependent on the value of the exponent, \( p \).
  • When \( p > 1 \), a p-series converges.
  • When \( p \leq 1 \), a p-series diverges.
These properties make p-series incredibly useful for the Limit Comparison Test, as they provide a benchmark for other series. In our exercise, the series \ \( \sum_{k=3}^{\infty} \frac{1}{(k-2)^{4}} \ \) was compared to the p-series with \( p = 4 \.\) This comparison was valid since \( p > 1 \,\) indicating convergence.
Convergent Series
A series is said to be convergent if the sum of its infinite terms approaches a finite number. This essentially means that as you add more and more terms of the series, the total tends to settle towards a particular value, not spiraling off into infinity or behaving erratically.
Convergence is crucial in understanding the overall behavior of a series. In applied mathematics and various fields like physics or engineering, convergent series often model real-world phenomena where infinite components contribute to a stable outcome. The conclusion of a series being convergent suggests a completeness in the summation process.
In mathematical terms, if a series \ \( \sum_{n=1}^{\infty} a_n \ \) has the property that \ \( \lim_{N \to \infty} S_N = L \ \,\) where \ \( S_N \ \,\) is the partial sum of the first \( N \ \) terms and \( L \ \,\) is a finite number, then the series is convergent. Using tools like the Limit Comparison Test helps identify such series.
Divergent Series
Unlike convergent series, divergent series do not settle on a single value or finite boundary as more terms are added. Instead, they might grow indefinitely, fluctuate without bound, or fail to settle to a particular value at all.
Divergence could indicate incomplete or chaotic summation, which does not converge to a meaningful or interpretable result. Identifying divergence is important, particularly in contexts where stability and predictability are required.
For a series like \ \( \sum_{n=1}^{\infty} a_n \,\) to be divergent, its partial sums \ \( S_N \ \,\) show any of these behaviors:
  • They tend toward infinity as \( N \) increases.
  • They oscillate or fluctuate without boundary.
  • They do not stabilize to any specific number.
In mathematical and scientific analyses, knowing divergence is just as critical as convergence, guiding whether or not a certain series can model or explain a situation accurately.

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Most popular questions from this chapter

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{c n}{b n+1}=\frac{c}{b}, \text { for real numbers } b > 0 \text { and } c > 0$$

Suppose that you take 200 mg of an antibiotic every 6 hr. The half-life of the drug is 6 hr (the time it takes for half of the drug to be eliminated from your blood). Use infinite series to find the long-term (steady-state) amount of antibiotic in your blood.

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} b^{-n}=0, \text { for } b > 1$$

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

Consider the sequence \(\left\\{x_{n}\right\\}\) defined for \(n=1,2,3, \ldots\) by $$x_{n}=\sum_{k=n+1}^{2 n} \frac{1}{k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2 n}.$$ a. Write out the terms \(x_{1}, x_{2}, x_{3}\). b. Show that \(\frac{1}{2} \leq x_{n}<1,\) for \(n=1,2,3, \ldots\). c. Show that \(x_{n}\) is the right Riemann sum for \(\int_{1}^{2} \frac{d x}{x}\) using \(n\) subintervals. d. Conclude that \(\lim _{n \rightarrow \infty} x_{n}=\ln 2\).

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