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Chapter 8: Problem 5

For what values of \(p\) does the series \(\sum_{k=10}^{\infty} \frac{1}{k^{p}}\) converge (initial index is 10 )? For what values of \(p\) does it diverge?

Short Answer

Expert verified
Answer: The series converges for values of \(p > 1\) and diverges for values of \(p \leq 1\).

Step by step solution

01

Understand the series type

The given series \(\sum_{k=10}^{\infty} \frac{1}{k^{p}}\) is a type of p-series, which is in the form of \(\sum_{k=1}^{\infty} \frac{1}{k^{p}}\). In this case, the initial index is given as 10, but this does not affect the convergence or divergence of the series.
02

Apply the p-series test

The p-series test states that the series \(\sum_{k=1}^{\infty} \frac{1}{k^{p}}\) converges if \(p > 1\) and diverges if \(p \leq 1\). In this case, we have the series \(\sum_{k=10}^{\infty} \frac{1}{k^{p}}\). The same rule applies due to having the same general form.
03

Determine the values of \(p\) for convergence and divergence

The series \(\sum_{k=10}^{\infty} \frac{1}{k^{p}}\) converges if \(p > 1\). Thus, for values of \(p\) where \(p > 1\), the series converges. On the other hand, the series diverges if \(p \leq 1\). That means for values of \(p\) where \(p \leq 1\), the series diverges.
04

Conclusion

The series \(\sum_{k=10}^{\infty} \frac{1}{k^{p}}\) converges for values of \(p > 1\) and diverges for values of \(p \leq 1\).

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Most popular questions from this chapter

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. $$a_{n}=n^{10} \text { and } b_{n}=n^{9} \ln ^{3} n, n \geq 7$$

Express each sequence \(\left\\{a_{n}\right\\}_{n=1}^{\infty}\) as an equivalent sequence of the form \(\left\\{b_{n}\right\\}_{n=3}^{\infty}\). $$\\{2 n+1\\}_{n=1}^{\infty}$$

Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \dots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots$$

A ball is thrown upward to a height of \(h_{0}\) meters. After each bounce, the ball rebounds to a fraction r of its previous height. Let \(h_{n}\) be the height after the nth bounce and let \(S_{n}\) be the total distance the ball has traveled at the moment of the nth bounce. a. Find the first four terms of the sequence \(\left\\{S_{n}\right\\}\) b. Make a table of 20 terms of the sequence \(\left\\{S_{n}\right\\}\) and determine \(a\) plausible value for the limit of \(\left\\{S_{n}\right\\}\) $$h_{0}=20, r=0.5$$

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.
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