Question

$\sum_{n=1}^{n=\infty} \frac{1}{(2n)^2}$

Answer 1

$\frac{\pi^2}{24}$

Answer 2

To evaluate the sum $\sum_{n=1}^{\infty} \frac{1}{(2n)^2}$:

1. Simplify the general term: The general term of the series is $\frac{1}{(2n)^2}$. Squaring the denominator gives $\frac{1}{4n^2}$.

2. Rewrite the summation: The series can be written as: $\sum_{n=1}^{\infty} \frac{1}{4n^2}$

3. Factor out the constant: The constant $\frac{1}{4}$ can be taken out of the summation: $\frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$

4. Recognize the known series (Basel Problem): The series $\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \dots$ is a famous series known as the Basel problem. Its sum is a well-known result in mathematics: $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

5. Substitute the value and calculate: Substitute the value of the Basel sum into our expression: $\frac{1}{4} \times \frac{\pi^2}{6} = \frac{\pi^2}{24}$

The sum of the series is $\frac{\pi^2}{24}$.