Question

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}.\) (Assume the result of Exercise 63.)

Short answer

Question: Prove that the sum of the reciprocals of the square of all positive integers, with alternating signs, is equal to \(\frac{\pi^2}{12}\). Answer: To prove that the sum of the reciprocals of the square of all positive integers, with alternating signs, is equal to \(\frac{\pi^2}{12}\), we used the given information that the sum of the reciprocals of the square of all positive integers is equal to \(\frac{\pi^2}{6}\). We subtracted the desired sum from the given sum, simplified the difference, and realized that the sum only has even terms. After factoring and substituting the given sum, we calculated the desired sum to be \(\frac{\pi^2}{12}\).

Step-by-step solution

Write down the given series and define the given sum

We are given that $$ \sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6}. $$ We are asked to calculate the sum $$ \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}. $$

Subtract the desired sum from the given sum

In order to find the sum of the required series, let's subtract it from the given sum. So, we write down the difference as follows: $$ \sum_{k=1}^{\infty}\frac{1}{k^2} - \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2} = \sum_{k=1}^{\infty}\left(\frac{1}{k^2}-\frac{(-1)^{k+1}}{k^2}\right). $$

Simplify the difference

Now, let's simplify the difference: $$ \sum_{k=1}^{\infty}\left(\frac{1}{k^2}-\frac{(-1)^{k+1}}{k^2}\right) = \sum_{k=1}^{\infty}\frac{1-(-1)^{k+1}}{k^2}. $$

Realize that the sum only has even terms

Notice that for odd values of k, \((-1)^{k+1} = 1\), while for even values of k, \((-1)^{k+1} = -1\). So, the sum reduces to: $$ \sum_{k=1}^{\infty}\frac{1-(-1)^{k+1}}{k^2} = \sum_{k=1}^{\infty}\frac{2}{(2k)^2}. $$ Now, the sum only contains even terms.

Factor 2 from the sum's numerator and 4 from the denominator

Now, let's factor 2 from the numerator and 4 from the denominator of the sum: $$ \sum_{k=1}^{\infty}\frac{2}{(2k)^2} = \sum_{k=1}^{\infty}\frac{1}{2}\cdot\frac{1}{k^2} = \frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k^2}. $$

Substitute the given sum

Now, let's substitute the value of the given sum \(\frac{\pi^2}{6}\) into the simplified difference: $$ \frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{1}{2}\cdot\frac{\pi^2}{6}. $$

Calculate the desired sum

Finally, we can calculate the desired sum: $$ \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2} = \sum_{k=1}^{\infty}\frac{1}{k^2} - \frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6} - \frac{1}{2}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{12}. $$ Therefore, $$ \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2} = \frac{\pi^2}{12}. $$

Key concepts

Convergence of series

When considering mathematical series, a key aspect we need to understand is convergence. A series converges if the partial sums of its terms approach a specific value. In simpler terms, as you add more and more terms of the series, the overall sum gets closer to a particular number. For instance, with the series \( \sum_{k=1}^{\infty} \frac{1}{k^2} \), it converges to \( \frac{\pi^2}{6} \). This means that even though there are infinitely many terms, their sum doesn’t grow indefinitely, but rather stabilizes at that value.

Convergence is essential when evaluating series because it informs us whether a series has a finite, useful total. If a series doesn't converge, it might suggest concepts or phenomena that can't be measured accurately or meaningfully with that series.

• To determine convergence, mathematicians employ tests such as the Ratio Test, Root Test, and others.
• Each test has its own conditions and can be applied based on the nature and behavior of the series' terms.

Series manipulation

Series manipulation involves transforming a series into a helpful or more manageable form. This is crucial for solving many mathematical problems, such as getting from one known sum to a related unknown. In our exercise, we start with an established convergence: \( \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} \).

The problem then asks us to handle an alternating series in a certain way. The alternating series \( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} \) can be derived by manipulating the original series. Here's how: subtract it from the given sum and observe how the terms change. We used: \( \sum_{k=1}^{\infty}\left(\frac{1}{k^2}-\frac{(-1)^{k+1}}{k^2}\right) \), which simplifies effectively.

• This manipulation reveals insights about the contribution of every term related to even or odd positioning in the series.
• By focusing on simplified results, as observed when even terms emerge more prominently, we gain a deeper understanding of the series' structure.

Understanding how to manipulate series through such transformations aids in exploring more complex mathematical constructs.

Infinite series

Infinite series are fascinating yet challenging because they involve an endless sum of terms. They are used to approximate values that cannot be exactly expressed otherwise. An infinite series can either converge to a specific value or diverge, meaning it grows infinitely large.

In the exercise, we observed two infinite series: one derived from straightforward terms \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) and another with alternating signs: \( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} \). While both are infinite, they display different characteristics due to the behavior of their terms.

• Infinite series often represent complex functions within mathematics, offering approximations for real-world applications.
• Tools such as Taylor Series and Fourier Series show the widespread utility of infinite series in various fields, including physics and engineering.

Mastering infinite series is a gateway to broader mathematical understanding and practical problem-solving, where capturing the essence of infinity with finite understanding is key.