Prove the following series

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In summary, the solution involves rewriting the given series in terms of an alternating series and using the known value of ##\frac{\pi^2}{6}## to obtain the correct answer of ##\frac{\pi^2}{8}-1##.
  • #1
JasonHathaway
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Homework Statement



Prove that, by putting x=0 x=∏ in [itex]x^{2}=\frac{\pi^{2}}{3}+4 \sum\limits_{n=1}^\infty \frac{1}{n^{2}} cos(nx) cos(n \pi)[/itex], that [itex]\frac{\pi^{2}}{8}= \sum\limits_{n=1}^\infty \frac{1}{(2n+1)^{2}}[/itex]

The Attempt at a Solution



1111.png


This a solved problem, I've understood the first two parts, and how the even elements of the series were eliminated, but what about [itex]\frac{\pi^{2}}{6} [/itex] and [itex]\frac{\pi^{2}}{8} [/itex]?
 
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  • #2
JasonHathaway said:

Homework Statement



Prove that, by putting x=0 x=∏ in [itex]x^{2}=\frac{\pi^{2}}{3}+4 \sum\limits_{n=1}^\infty \frac{1}{n^{2}} cos(nx) cos(n \pi)[/itex], that [itex]\frac{\pi^{2}}{8}= \sum\limits_{n=1}^\infty \frac{1}{(2n+1)^{2}}[/itex]


Homework Equations





The Attempt at a Solution



<image>

This a solved problem, I've understood the first two parts, and how the even elements of the series were eliminated, but what about [itex]\frac{\pi^{2}}{6} [/itex] and [itex]\frac{\pi^{2}}{8} [/itex]?

The answer is actually incorrect. The correct answer is ##\frac{\pi^2}{8}-1##.

Rewrite series 2 in the following way:
$$\frac{\pi^2}{3}=4\left(\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots \right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots \right)\right)$$
$$\Rightarrow \frac{\pi^2}{3}=4\left(1+\sum_{n=1}^{\infty} \frac{1}{(2n+1)^2} -\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots \right)\right)\,\,\,\,\,\,(*)$$
From (3), you have ##\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots##, put this in ##(*)## and you should be able to reach the correct answer.
 
  • #3
You can turn the equation (2) into an expression for ##\frac{\pi^2}{12}## easily enough, and then ##\frac{\pi^2}{12}+\frac{\pi^2}{6}=\dots##

(and I disagree with Pranav-Arora - ##\frac{\pi^2}{8}## is correct)

Edit - Ah I see now what Pranav-Arora meant - the final summation should run from n=0, otherwise, indeed, the LHS should be ##\frac{\pi^2}{8}-1##
 
Last edited:

What does it mean to "prove" a series?

Proving a series involves demonstrating that the series converges or diverges. This can be done through various methods, such as the comparison test, the ratio test, or the integral test.

What is a series?

A series is a sum of infinitely many terms. It is represented in sigma notation as Σ(an), where "a" represents the terms of the series and "n" represents the index of the terms.

What are the different types of series?

There are several types of series, including arithmetic series, geometric series, telescoping series, and power series. Each type has its own specific formula for calculating the sum.

What is the importance of proving a series?

Proving a series is important because it allows us to determine whether the series converges or diverges. This information can be used to evaluate the sum of a series and to make predictions about the behavior of a function.

Can all series be proven?

No, not all series can be proven. Some series are too complex to be evaluated using traditional methods and may require advanced techniques such as the Cauchy condensation test or the alternating series test.

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