Question

Reciprocals of odd squares Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\)

Short answer

Answer: The sum of the reciprocals of odd squares is \(\frac{3\pi^2}{24}\).

Step-by-step solution

Identify the given information

We are given that the sum of the reciprocals of all squares is: $$\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$$ We are asked to find the sum of the reciprocals of odd squares, so we will represent this as: $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$$

Express the sum of the reciprocals of odd squares in terms of all squares

In order to relate the sum of the reciprocals of all squares and the sum of the reciprocals of odd squares, we will subtract the sum of the reciprocals of even squares from the sum of the reciprocals of all squares. Let's represent the sum of the reciprocals of even squares as: $$\sum_{n=1}^{\infty} \frac{1}{(2n)^2}$$ Now, subtracting the sum of the reciprocals of even squares from the sum of the reciprocals of all squares, we have: $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \sum_{k=1}^{\infty} \frac{1}{k^2} - \sum_{n=1}^{\infty} \frac{1}{(2n)^2}$$

Simplify the equation and find the sum of the reciprocals of odd squares

Now, let's simplify the equation from step 2: $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \sum_{n=1}^{\infty} \frac{1}{(2n)^2}$$ $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \sum_{n=1}^{\infty} \frac{1}{4n^2}$$ $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2}$$ Now, we can replace the sum of the reciprocals of all squares with given value \(\frac{\pi^2}{6}\): $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{1}{4}\left(\frac{\pi^2}{6}\right)$$ Finally, to find the sum of the reciprocals of odd squares, we can simplify the equation: $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6}\left(1 - \frac{1}{4}\right)$$ $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6}\left(\frac{3}{4}\right)$$ $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{3\pi^2}{24}$$

Key concepts

Reciprocal Sums

Understanding reciprocal sums is key to tackling infinite series problems. Essentially, a reciprocal sum is a series where each term is the reciprocal (or, one divided by) of the elements in a particular sequence.
For instance, in an infinite series involving the reciprocals of squares, each term is based on the square of each whole number. Like this:

• The reciprocal of 1 is \( \frac{1}{1} \)
• The reciprocal of 4 is \( \frac{1}{4} \)
• The reciprocal of 9 is \( \frac{1}{9} \)

When summed, these reciprocals create an infinite series. Identifying how these terms add up is important to solve various mathematical problems. For example, the famous Basel problem's solution is the sum of the reciprocals of all squares — \( \frac{\pi^2}{6} \). Such kinds of reciprocal sums reveal fascinating truths and relationships in mathematics!

Odd Squares

Odd squares are the squares of odd numbers, like 1, 9, 25, and so forth. These numbers come from multiplying an odd number by itself.

• \[ 1 \times 1 = 1 \]
• \[ 3 \times 3 = 9 \]
• \[ 5 \times 5 = 25 \]

Reciprocal sums involving odd squares only consider the reciprocals of numbers like those listed above. This means we exclude squares of even numbers entirely from our calculations.
In the given exercise, understanding the sum of reciprocal odd squares involves subtracting the contribution of even squares from the total sum of squares. So, if we know the sum of reciprocals of all squares is \( \frac{\pi^2}{6} \), and we find the sum of even squares' reciprocals, subtracting gives us the sum for odd squares.
It's like filtering out even squares to focus on the effects of odd ones in the whole picture of reciprocal sums.

Even Squares

Even squares result from squaring an even number. Examples include 4, 16, and 36, among others.

• \[ 2 \times 2 = 4 \]
• \[ 4 \times 4 = 16 \]
• \[ 6 \times 6 = 36 \]

These even squares are significant when distinguishing between reciprocal sums of all numbers, odd squares, and even squares. By understanding the contribution of even squares, we can compute the sum of odd square reciprocals more precisely.
In the exercise, recognizing the sum of reciprocal even squares is crucial. When \( \sum_{n=1}^{\infty} \frac{1}{(2n)^2} \) is taken and subtracted from the total, it leaves us with only the odd square contributions.
This logical breakdown not only simplifies complex series problems but also equips students with a methodical approach for similar mathematical challenges in the future.