Chapter 7: Q15P (page 358)

Use Problem 5.7to show that $\frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \frac{1}{6^{2} - 1} + \dots = \frac{1}{2}$

Short Answer

Expert verified

The resultant expansion is $\sum_{n = 2 k}^{\infty} \frac{1}{n^{2} - 1} = \frac{1}{2} k = 1, 2, 3, 4$ .

Step by step solution

Given data

The given function is $\frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \frac{1}{6^{2} - 1} + \dots = \frac{1}{2}$ .

Concept of Dirichlet’s theorem

Dirichlet's theorem states that the collection of all numbers of the form $n a + b$ contains an unlimited number of prime numbers, where the constants a and b are integers with no common divisors other than 1 and the variable nis any natural number.

Simplify the expression

Use the Problem (5.11) with $x = 0$ .

$f (0) = 0 f (0) = \frac{1}{π} + 0 - \frac{2}{π} \sum_{n = 2 k}^{\infty} \frac{1}{n^{2} - 1} f (0) = 0 \sum_{n = 2 k}^{\infty} \frac{1}{n^{2} - 1} f (0) = \frac{1}{2}$

Therefore, expansion is $\sum_{n = 2 k}^{\infty} \frac{1}{n^{2} - 1} = \frac{1}{2} k = 1, 2, 3, 4$ .

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Use Problem 5.7to show that $\frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \frac{1}{6^{2} - 1} + \dots = \frac{1}{2}$

Short Answer

Step by step solution

Given data

Concept of Dirichlet’s theorem

Simplify the expression

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Short Answer

Step by step solution

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Simplify the expression

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Use Problem 5.7to show that $\frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \frac{1}{6^{2} - 1} + \dots = \frac{1}{2}$