How Do You Solve the Alternating Series in POTW #283?

  • MHB
  • Thread starter Euge
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    2017
In summary, POTW #283 is a weekly problem-solving challenge posted by online communities and organizations. Its purpose is to encourage critical thinking and problem-solving skills in the fields of science, technology, engineering, and mathematics. The solution to POTW #283 varies and requires a unique and creative approach. To participate, you can search for the problem online or join online communities. The benefits of participating include improving critical thinking, problem-solving, and teamwork skills, as well as learning from others and discovering new approaches to problem-solving.
  • #1
Euge
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Ackbach has asked me to step in for him for a while. Here is this week's POTW:

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Evaluate the sum of the alternating series $$\sum\limits_{n = 1}^\infty \frac{(-1)^{n-1}}{n^4}$$
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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
Congratulations to Opalg for his correct solution. You can read his solution below.
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4} = \sum_{n=1}^\infty\frac1{n^4} - 2\sum_{n=1}^\infty\frac1{(2n)^4} = \Bigl(1 - \frac2{16}\Bigr)\sum_{n=1}^\infty\frac1{n^4} = \frac{7\pi^4}{720},$$ using the famous fact that \(\displaystyle \sum_{n=1}^\infty\frac1{n^4} = \zeta(4) = \frac{\pi^4}{90}.\)

One way to prove the formula \(\displaystyle \sum_{n=1}^\infty\frac1{n^4} = \frac{\pi^4}{90}\) is to find the Fourier series for the function $f(x) = x^2$ on the interval $[-\pi,\pi].$

For $n\ne0$, $$\hat{\ f}(n) = \frac1{2\pi}\int_{-\pi}^\pi x^2e^{inx}dx = \frac1{2n\pi i}\Bigl[x^2e^{inx}\Bigr]_{-\pi}^\pi - \frac1{n\pi i}\int_{-\pi}^\pi xe^{inx}dx = 0 + \frac1{n^2\pi}\Bigl[xe^{inx}\Bigr]_{-\pi}^\pi - \frac1{n^2\pi}\int_{-\pi}^\pi e^{inx}dx = \frac{2(-1)^n}{n^2}$$ (integrating by parts twice). For $n=0$, $$\hat{\ f}(0) = \frac1{2\pi}\int_{-\pi}^\pi x^2dx = \frac1{2\pi}\Bigl[\frac{x^3}3\Bigr]_{-\pi}^\pi = \frac{\pi^2}3.$$

Now apply Parseval's theorem, which says that \(\displaystyle \|f\|_2^2 = \sum_{n\in\Bbb{Z}}|\hat{\ f}(n)|^2\), where $$\|f\|_2^2 = \frac1{2\pi}\int_{-\pi}^\pi|f(x)|^2dx = \frac1{2\pi}\int_{-\pi}^\pi x^4 dx = \frac{\pi^4}5.$$ That gives \(\displaystyle \frac{\pi^4}5 = \frac{\pi^4}9 + 2\sum_{n=1}^\infty\frac4{n^4}\), from which \(\displaystyle \sum_{n=1}^\infty\frac1{n^4} = \frac{\pi^4}{90}.\)
 

FAQ: How Do You Solve the Alternating Series in POTW #283?

What is POTW #283?

POTW #283 stands for "Problem of the Week #283". It is a weekly problem-solving challenge posted by various online communities and organizations.

What is the purpose of POTW #283?

The purpose of POTW #283 is to encourage critical thinking and problem-solving skills in individuals, particularly in the fields of science, technology, engineering, and mathematics (STEM).

What is the solution to POTW #283?

The solution to POTW #283 varies depending on the specific problem given. It is typically a unique and creative approach to solving the given problem.

How can I participate in POTW #283?

To participate in POTW #283, you can search for the problem on online platforms or subscribe to organizations that post weekly challenges. You can also join online communities that share solutions and discuss strategies for solving the problems.

What are the benefits of participating in POTW #283?

Participating in POTW #283 can improve your critical thinking, problem-solving, and teamwork skills. It also allows you to learn from others and discover new approaches to solving problems.

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