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Chapter 9: Problem 57

Choose your test Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{k^{8}}{k^{11}+3}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Identify a simpler series to compare with the given series

The given series is: $$\sum_{k=1}^{\infty} \frac{k^{8}}{k^{11}+3}$$ Let's compare it with the simpler series: $$\sum_{k=1}^{\infty} \frac{1}{k^{3}}$$ We will now use the limit comparison test.
02

Apply the Limit Comparison Test

According to the limit comparison test, if the limit of the ratio of the given series' terms to the simpler series' terms exists and is a positive number, then both series either converge or diverge together. Therefore, we now compute the limit: $$\lim_{k \to \infty} \frac{\frac{k^{8}}{k^{11}+3}}{\frac{1}{k^{3}}}$$
03

Simplify the limit expression

To simplify the above limit, multiply the numerator and denominator by \(k^{11}\): $$\lim_{k \to \infty} \frac{k^{8}}{k^{11}+3} \cdot \frac{k^{11}}{1} = \lim_{k \to \infty} \frac{k^{19}}{k^{11}+3}$$ Now, divide both numerator and denominator by \(k^{19}\): $$\lim_{k \to \infty} \frac{k^{19}}{k^{11}+3} \cdot \frac{\frac{1}{k^{19}}}{\frac{1}{k^{19}}} = \lim_{k \to \infty} \frac{1}{\frac{k^{11}}{k^{19}}+\frac{3}{k^{19}}}$$
04

Compute the limit

As \(k\) approaches infinity, the limit of the denominator becomes: $$\lim_{k \to \infty} \frac{k^{11}}{k^{19}} + \lim_{k \to \infty} \frac{3}{k^{19}} = 0 + 0 = 0$$ So the limit of the whole expression is: $$\lim_{k \to \infty} \frac{1}{0} = \infty$$ Since the limit is infinite, the limit comparison test is inconclusive.
05

Use the Comparison Test

Since the limit comparison test was inconclusive, let's try the comparison test. We only need to show that the given series is smaller than the simpler series we've chosen, which is the p-series with p = 3: $$\sum_{k=1}^{\infty} \frac{1}{k^{3}}$$ Notice that for all \(k \geq 1\): $$ \frac{k^{8}}{k^{11}+3} \leq \frac{k^{8}}{k^{11}} = \frac{1}{k^{3}} $$ As we know that the p-series with \(p=3\) converges, by the comparison test, the given series also converges: $$\sum_{k=1}^{\infty} \frac{k^{8}}{k^{11}+3}$$

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Most popular questions from this chapter

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd. } \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N,\) the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7,\) and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5 7, \(11,13, \ldots .\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=1}^{\infty} \ln \left(\frac{k}{k+1}\right)^{p}$$

Evaluate the limit of the following sequences. $$a_{n}=\int_{1}^{n} x^{-2} d x$$

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(R_{n}<10^{-6}\) ). a. \(\sum_{k=0}^{\infty} 0.6^{k}\) b. \(\sum_{k=0}^{\infty} 0.15^{k}\)
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