/*! 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 60 Alternating \(p\) -series Given ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Alternating \(p\) -series Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90},\) show that \(\left.\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}}=\frac{7 \pi^{4}}{720} . \text { (Assume the result of Exercise } 63 .\right)\)

Short Answer

Expert verified
Question: Show that the sum of the alternating p-series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} = \frac{7\pi^4}{720}\) given the sum of the p-series \(\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}\). Answer: \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} = \frac{7\pi^4}{720}\)

Step by step solution

01

Find the sum of even terms and odd terms separately

Start by breaking the alternating p-series sum into two parts - the sum of even terms (positive) and the sum of odd terms (negative): \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} = \sum_{k=1}^{\infty} \frac{1}{(2k)^4} - \sum_{k=1}^{\infty} \frac{1}{(2k-1)^4}\).
02

Simplify the even and odd terms sum

Next, simplify the even and odd terms sum: \(\sum_{k=1}^{\infty} \frac{1}{(2k)^4} - \sum_{k=1}^{\infty} \frac{1}{(2k-1)^4} = \frac{1}{16}\sum_{k=1}^{\infty} \frac{1}{k^4} - \sum_{k=1}^{\infty} \frac{1}{(2k-1)^4}\).
03

Use the given p-series sum and rewrite the equation

Use the given p-series sum, \(\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}\), to rewrite the equation: \(\frac{1}{16}\cdot \frac{\pi^4}{90} - \sum_{k=1}^{\infty} \frac{1}{(2k-1)^4} = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4}\).
04

Solve for the sum of odd terms

Rearrange the equation to solve for the sum of odd terms: \(\sum_{k=1}^{\infty} \frac{1}{(2k-1)^4} = \frac{\pi^4}{90} - \frac{\pi^4}{16 \cdot 90}\).
05

Combine the two fractions

Combine the two fractions on the right side of the equation: \(\sum_{k=1}^{\infty} \frac{1}{(2k-1)^4} = \frac{(16-1)\pi^4}{16 \cdot 90} = \frac{15\pi^4}{16 \cdot 90}\).
06

Plug the expression back into the equation

Plug the expression for the sum of odd terms back into the equation for the alternating p-series sum, then simplify: \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} = \frac{\pi^4}{16 \cdot 90} - \frac{15\pi^4}{16 \cdot 90} = \frac{\pi^4}{16 \cdot 90} \cdot (1 - 15) = \frac{-14\pi^4}{16 \cdot 90}\).
07

Simplify the fraction

Finally, simplify the fraction: \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} = \frac{7\pi^4}{8 \cdot 90} = \frac{7\pi^4}{720}\). Thus, we have shown that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} = \frac{7\pi^4}{720}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

P-Series
A p-series is a specific type of mathematical series that takes the form \[\sum_{k=1}^{\infty} \frac{1}{k^p},\]where \( p \) is a constant. The behavior and characteristics of a p-series largely depend on the value of \( p \).
  • When \( p \gt 1 \), the p-series is known to be convergent, meaning it adds up to a finite sum. This is because the terms in the series diminish rapidly enough to sum to a limit.
  • If \( p \leq 1 \), the p-series is divergent, indicating that the series does not sum to a finite number. The terms decrease too slowly, and the series infinitely grows.
Understanding the nature of p-series helps in analyzing more complex series, especially when determining convergence or divergence modes. The knowledge of p-series provides a foundational tool in the broader analysis of series patterns.
Convergence
Convergence in mathematics generally refers to the idea that as you sum more and more terms of a series, it tends to approach a certain finite value. In the context of series, it’s crucial because it expresses whether the series has a limiting value.
For alternating series especially, convergence can often be determined by the Alternating Series Test. This test states that if the absolute value of the terms decreases steadily to zero, the series converges.
For example, in the \[\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4}\]alternating series we were analyzing, not only the p-series rules apply, but the convergence hinges on the fact that the series alternates and diminishes successfully, allowing it to sum to the fixed value \( \frac{7\pi^4}{720} \). Understanding the mechanics of convergence is vital for series manipulation and mathematical analysis.
Series Manipulation
Series manipulation entails various techniques used to rewrite or rearrange a series to explore its properties more clearly or simplify calculations.
In our exercise example, the series manipulation involved splitting the alternating p-series into even and odd terms. This methodial breakdown allows easier computation:
  • By expressing the series as \[\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} = \sum_{k=1}^{\infty} \frac{1}{(2k)^4} - \sum_{k=1}^{\infty} \frac{1}{(2k-1)^4}\]we derived a relationship between components.
  • This facilitated simplifying and substituting known values like \[\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}\].
Such manipulation is a powerful tool. It often reveals insights or solves parts of series-related problems efficiently, making complex series manageable.
Mathematical Proofs
A mathematical proof is a logical argument that establishes the truth of a given statement or equation.
Proofs form the backbone of mathematics, ensuring that conjectures become accepted truths. In this exercise, the process of proving the alternating p-series value \[\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} = \frac{7\pi^4}{720}\]involves a meticulously structured sequence:
  • Starting with a known p-series sum, \[\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}\], provides a basis.
  • Addition and subtraction were employed within series components (even and odd), leveraging manipulation techniques.
  • Meticulous calculation and combination followed, leading to the result by performing careful arithmetic.
Understanding proofs involves not just the final result but comprehending the logical steps taken. Each step must be clear, with justification for actions taken, ensuring accuracy and truth in mathematical conclusions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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?

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$

The famous Fibonacci sequence was proposed by Leonardo Pisano, also known as Fibonacci, in about \(\mathrm{A.D.} 1200\) as a model for the growth of rabbit populations. It is given by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1 .\) Each term of the sequence is the sum of its two predecessors. a. Write out the first ten terms of the sequence. b. Is the sequence bounded? c. Estimate or determine \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}},\) the ratio of the successive terms of the sequence. Provide evidence that \(\varphi=(1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Use induction to verify the remarkable result that $$f_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n}-(-1)^{n} \varphi^{-n}\right).$$

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

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. $$a_{n}=\sqrt{n} \text { and } b_{n}=2 \ln n, n \geq 3$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.