Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
Answer: \(\frac{7 \pi^{4}}{720}\)

Step by step solution

01

Write down the given series and the desired series

We are given the series: $$\sum_{k=1}^{\infty} \frac{1}{k^{4}} = \frac{\pi^{4}}{90}$$ And we aim to find: $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}} = S$$
02

Add the given series to the desired series

In order to manipulate the desired series, add the given series to it: $$\sum_{k=1}^{\infty} \frac{1}{k^{4}} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}} = \frac{\pi^{4}}{90} + S$$ $$(2) \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}} = \frac{\pi^{4}}{90} + S$$
03

Rearrange the equation to find the desired series

Now, solve the equation for \(S\), which represents the desired series: $$S = \frac{(2) \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}} - \frac{\pi^{4}}{90}}{3} = \frac{\pi^{4}}{180} - \frac{\pi^{4}}{90} = \frac{\pi^{4}}{60} - \frac{\pi^{4}}{30} = \frac{7 \pi^{4}}{720}$$ 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 Vaia!

Key Concepts

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

Riemann Zeta Function
The Riemann Zeta Function, denoted by \( \zeta(s) \), is a fundamental tool in number theory and complex analysis, defined for complex numbers \( s \) with a real part greater than 1. This function is an infinite series represented as:
  • \( \zeta(s) = \sum_{k=1}^{\infty} \frac{1}{k^{s}} \)
Here, each term contributes the reciprocal of \( k \) raised to the power of \( s \). For special values, such as \( s = 2, 4, 6, \ldots \), the series converges to well-known constants, including expressions involving \( \pi \), like \( \zeta(4) = \frac{\pi^4}{90} \).
The Riemann Zeta Function extends beyond real numbers and is central to the distribution of prime numbers. For values where the real part of \( s \) is less than or equal to 1, it is analytically continued, except at \( s = 1 \), where it diverges. When we deal with alternating series, like in the given problem, the zeta function can be adjusted by transformation principles that account for alternating signs.
Convergence of Series
Convergence of an infinite series is a crucial concept in analysis. For a series to converge, the sum of its infinite terms must approach a specific limit. This entails the partial sums of the series getting closer to a finite value as more terms are added.
Consider the series \( \sum_{k=1}^{\infty} \frac{1}{k^{4}} \) from our example. This series is known to converge, and its sum is given as \( \frac{\pi^4}{90} \). Its convergence is assured since the terms \( \frac{1}{k^4} \) become very small rapidly due to the fourth power in the denominator. The convergence is guaranteed when the exponent in the series \( \frac{1}{k^p} \) is greater than 1, known as the p-series test.
  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
In alternating series like \( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}} \), Liebnitz's test is commonly used to ensure convergence. The conditions include that the terms' magnitude decreases and approaches zero as \( k \) increases.
Infinite Series
Infinite series can be seen in mathematics where a sequence extends indefinitely. Understanding this concept involves evaluating the behavior and properties of summing an endless number of terms. For example, in our exercise, both series involved have an infinite number of terms, each term given by \( \frac{1}{k^{4}} \), but with different signs.
Infinite series can be:
  • Convergent - where they sum to a finite limit, like our harmonic series in the example.
  • Divergent - where no finite limit is approached as more terms are added.
Exploring the realm of infinite series aids in understanding various mathematical and physical phenomena, including geometric series, power series, and even Fourier series in signal processing. In applied mathematics, series are useful for approximating functions and solving complex problems that cannot be addressed by finite processes. In our problem, by considering an infinite alternating series, it allows us to calculate the adjusted sum using the known values of the regular infinite series.

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

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{\cos k}{k^{3}}$$

The CORDIC (COordinate Rotation DIgital Calculation) algorithm is used by most calculators to evaluate trigonometric and logarithmic functions. An important number in the CORDIC algorithm, called the aggregate constant, is \(\prod_{n=0}^{\infty} \frac{2^{n}}{\sqrt{1+2^{2 n}}},\) where \(\prod_{n=0}^{N} a_{n}\) represents the product \(a_{0} \cdot a_{1} \cdots a_{N}\). This infinite product is the limit of the sequence $$\left\\{\prod_{n=0}^{0} \frac{2^{n}}{\sqrt{1+2^{2 n}}} \cdot \prod_{n=0}^{1} \frac{2^{n}}{\sqrt{1+2^{2 n}}}, \prod_{n=0}^{2} \frac{2^{n}}{\sqrt{1+2^{2 n}}} \ldots .\right\\}.$$ Estimate the value of the aggregate constant.

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$

Evaluate the limit of the following sequences. $$a_{n}=\frac{7^{n}}{n^{7} 5^{n}}$$

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 $$ \ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln n $$ 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, but not proved, 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 \(10 ?\)

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.

Sign-up for free